Asymptotics for evolution problems with nonlocal

Asymptotics for evolution problems with nonlocal diffusion Julio D. Rossi IMDEA Matemáticas, C-IX, Campus Cantoblanco, UAM, Madrid, Spain and Departamento de Matemática, FCEyN UBA (1428) Buenos Aires, Argentina. [email protected] http://mate.dm.uba.ar/∼jrossi

Abstract. In these notes we review recent results concerning solutions to nonlocal evolution equations with different boundary conditions, Dirichlet or Neumann and even for the Cauchy problem. We deal with existence/uniqueness of solutions and their asymptotic behavior. We also review some recent results concerning limits of solutions to nonlocal equations when a rescaling parameter goes to zero. We recover in these limits some of the most frequently used diffusion models: the heat equation with Neumann or Dirichlet boundary conditions, the p−Laplace evolution equation with Neumann boundary conditions and a convection-diffusion equation.

3

Acknowledgements To Mayte Perez-LLanos, for her patience and encouragement. The author wants to thank to F. Andreu, J. Mazon, J. Toledo, L. Ignat, C. Schönlieb, C. Cortazar and M. Elgueta for their enthusiasm and friendship working together.

CHAPTER 1

Introduction

11

Dir12.intro

First, let us briefly introduce the prototype of nonlocal problem that will be considered along this work. R Let J : RN → R be a nonnegative, radial, continuous function with RN J(z) dz = 1. Nonlocal evolution equations of the form Z (1.1) ut (x, t) = (J ∗ u − u)(x, t) = J(x − y)u(y, t) dy − u(x, t), RN

and variations of it, have been recently widely used to model diffusion processes. More F precisely, as stated in [59], if u(x, t) is thought of as a density at the point x at time t and J(x − y) is thought of as the probability distribution of jumping from location y to R location x, then RN J(y − x)u(y, t) dy = (J ∗ u)(x, t) is the rate R at which individuals are arriving at position x from all other places and −u(x, t) = − RN J(y − x)u(x, t) dy is the rate at which they are leaving location x to travel to all other sites. This consideration, in the absence of external or internal sources, leads immediately to the fact that the density 11 BCh BCh2 BFRW u satisfies equation ( 1.1). For recent references on nonlocal diffusion see, [15], [16], [17], CF C F [32], [35], [59], and references therein. These type of problems have been used to modelKOJ veryGOdifferent applied situations, CF MogEdel BodVel for example in biology ( [32], [71]), image processing ( [70], [63]), particle systems ( [22]), FourLau coagulation models ([61]), etc. Concerning boundary conditions for nonlocal problems we consider a bounded smooth domain Ω ⊂ RN and look at the nonlocal problem Z ut (x, t) = x ∈ Ω, t > 0, J(x − y)u(y, t) dy − u(x, t), (1.2)

RN

u(x, t) = 0,

x 6∈ Ω, t > 0,

u(x, 0) = u0 (x),

x ∈ Ω.

In this model we have that diffusion takes place in the whole RN but we impose that u vanishes outside Ω. This is the analogous of what is called Dirichlet boundary conditions for the heat equation. However, the boundary data is not understood in the usual sense, since we are not imposing that u|∂Ω = 0. 5

1.Neu.intro

6

1. INTRODUCTION

Let us turn our attention to Neumann boundary conditions. We study Z (1.3)

ut (x, t) =

J(x − y)(u(y, t) − u(x, t)) dy,

x ∈ Ω, t > 0,

Ω

u(x, 0) = u0 (x),

x ∈ Ω.

In this model we have that the integral Rterms take into account the diffusion inside Ω. In fact, as we have explained the integral J(x − y)(u(y, t) − u(x, t)) dy takes into account the individuals arriving or leaving position x from other places. Since we are integrating in Ω, we are imposing that diffusion takes place only in Ω. The individuals may not enter nor leave Ω. This is the analogous of what is called homogeneous Neumann boundary conditions in the literature. Here we review some results concerning existence and uniqueness for these models and their asymptotic behavior as t → ∞. These results says that these problems are well posed in appropriate functional spaces although they do not have a smoothing property. Moreover, the asymptotic behavior for the linear nonlocal models coincide with the one that holds for the heat equation. We will also review some recent results concerning limits of nonlocal problems when a scaling parameter (that measures the radius of influence of the nonlocal term) goes to zero. We recover in these limits some well known diffusion problems, namely, the heat equation with Neumann or Dirichlet boundary conditions, the p−Laplace equation with Neumann boundary conditions and a convection-diffusion equation. The content AMRT of these notes summarizes the research of the author in the last years and AMTR-sandpiles AMRTp-Lap ChChR CER CER2 CERW CERW2 IgR liv.rossi RoSho is contained in [7], [8], [9], [34], [40], [41], [42], [43], [67], [68], [76]. We refer to these papers for extra details and further references. There is a huge amount of papers dealing with nonlocal problems. Among them we BCC BFRW CC C Co1 Co2 CD1 CD2 Sch Z quote [14], [17], [38], [35], [44], [45], [47], [48], [81] andBC[87],CR1 devoted to travelling front CR2 CDM PR type solutions to the parabolic problem in Ω = R, and [19], [36], [37], [46], [75], which dealt with source term of logistic type, bistable or power-like nonlinearity. The particular ChChR liv.rossi instance of the parabolic problem in RN is considered in [34], [68], while the “Neumann” AMRTCERW CERW2 IgR boundary condition for the same problemCCEM is treated in [7], [42] and [43]. See also [67] for CER the appearance of convective terms and [39], [40] HMMV for interesting features in other related nonlocal problems. We finally mention the paper [66], where some logistic equations and systems of Lotka-Volterra type are studied. There is also an increasing interest in freecaffaSilSal boundary problems and regularity issues for BarIm caffaSil sil nonlocal problems. We refer to [12], [30], [31], [82], but we are not dealing with such issues in the present work. The Bibliography of this work does not escape the usual rule of being incomplete. In general, we have listed those papers which are more close to the topics discussed here. But, even for those papers, the list is far from being exhaustive and we apologize for omissions.

1. INTRODUCTION

7

These notes are organized as follows: sec.lineal

(1) in Chapter 2 we deal with the linear diffusion equation in the whole RN or in a bounded domain with Dirichlet or Neumann boundary condition paying special attention to the asymptotic behavior as t → ∞; sec.linear.refined

(2) in Chapter 3 we find refined asymptotics; sec.linear.higher

(3) in Chapter 4 we deal with higher order nonlocal problems; sec.linear.neu

(4) in Chapter 5 we approximate the heat equation with Neumann boundary conditions; sec.linear.Dirichlet

(5) in Chapter 6 we approximate the heat equation with Dirichlet boundary conditions; sec.scaling.higher

(6) in Chapter 7 we deal with approximations of higher order problems; sec.convection.diffsuion

(7) in Chapter 8 we present a nonlocal convection-diffusion equation; sec.nonlin.Neu

(8) in Chapter 9 we deal with a nonlinear nonlocal Neumann problem. In this Chapter we can consider a nonlocal analogous to the porous medium equation; sec.p.lapla.Neu

(9) in Chapter 10 we face a nonlocal diffusion model analogous to the p−Laplacian with Neumann boundary conditions, sec.sandpiles

(10) finally, in Chapter 11 we take the limit as p → ∞ in the previous model.

CHAPTER 2

The linear problem sec.lineal

The aim of this chapter is to study the asymptotic behavior of solutions of a nonlocal diffusion operator in the whole RN or in a bounded smooth domain with Dirichlet or Neumann boundary conditions.

11.2

0.1. The Cauchy problem. We will consider the linear nonlocal diffusion problem presented in the Introduction Z ut (x, t) = J ∗ u − u(x, t) = J(x − y)u(y, t) dy − u(x, t), (2.1) RN u(x, 0) = u0 (x), 11.2

We will understand a solution of (2.1) as a function 11.2

u ∈ C 0 ([0, +∞); L1 (RN ))

existencia.unicidad.cauchy

that verifies (2.1) in the integral sense, see Theorem 3. Our first result states that the decay rate as t goes to infinity of solutions of this nonlocal problem is determined by the behavior of the Fourier transform of J near the origin. The asymptotic decays are the same as the ones that hold for solutions of the evolution problem with right hand side given by a power of the laplacian. In the sequel we denote by fˆ the Fourier transform of f . Let us recall our hypotheses on J that we will assume throughout this chapter, R (H) J ∈ C(RN , R) is a nonnegative, radial function with RN J(x) dx = 1. ˆ This means that J is a radial density probability which implies obviously that |J(ξ)| ≤1 ˆ = 1, and we shall assume that Jˆ has an expansion of the form with J(0) ˆ = 1 − A|ξ|α + o(|ξ|α ) J(ξ) for ξ → 0 (A > 0). Remark that in this case, (H) implies also that 0 < α ≤ 2 and α 6= 1 if J has a first momentum. The main result of this chapter reads as follows, casita

asymp.hat.J

11.2

Theorem 1. Let u be a solution of (2.1) with u0 , uˆ0 ∈ L1 (RN ). If there exist A > 0 and 0 < α ≤ 2 such that ˆ = 1 − A|ξ|α + o(|ξ|α ), (2.2) J(ξ) ξ → 0, 9

10

2. THE LINEAR PROBLEM

then the asymptotic behavior of u(x, t) is given by lim tN/α max |u(x, t) − v(x, t)| = 0,

t→+∞

x

where v is the solution of vt (x, t) = −A(−∆)α/2 v(x, t) with initial condition v(x, 0) = u0 (x). Moreover, we have ku(·, t)kL∞ (RN ) ≤ C t−N/α , and the asymptotic profile is given by ¯ ¯ lim max ¯tN/α u(yt1/α , t) − ku0 kL1 GA (y)¯ = 0, t→+∞

y

α where GA (y) satisfies GÂ (ξ) = e−A|ξ| .

In the special case α = 2, the decay rate is t−N/2 and the asymptotic profile is a gaussian ˆ GA (y) = (4πA)N/2 exp(−A|y|2 /4) with A · Id = −(1/2)D2 J(0). Note that in this case (that occurs, for example, when J is compactly supported) the asymptotic behavior is the same as the one for solutions of the heat equation and, as happens for the heat equation, the asymptotic profile is a gaussian. The decay in L∞ of the solutions together with the conservation of mass give the decay of the Lp -norms by interpolation. As a consequence of the previous theorem, we find that this decay is analogous to the decay of the evolution given by the fractional laplacian, that is, ¡ ¢ 1− p1 −N α , ku(·, t)kLp (RN ) ≤ C t teo.decay.Lp

CC

see Corollary 11. We refer to [38] for the decay of the Lp -norms for the fractional laplacian, CH DZ EZ see also [33], [52] and [54] for finer decay estimates of Lp -norms for solutions of the heat equation. We shall make an extensive use of the Fourier transform in order to obtain explicit solutions in frequency formulation. Let us recall that if f ∈ L1 (RN ) then fˆ and fˇ are bounded and continuous, where fˆ is the Fourier transform of f and fˇ its inverse Fourier transform. Moreover, lim fˆ(ξ) = 0

|ξ|→∞

lim fˇ(x) = 0.

and

|x|→∞

We begin by collecting some properties of the function J. lem.J

Lemma 2. Let J satisfy hypotheses (H). Then, ˆ ˆ i) |J(ξ)| R ≤ 1, J(0) = 1. ii) If RN J(x)|x| dx < +∞ then Z ³ ´ ∇ξ Jˆ (0) = −i i

xi J(x) dx = 0 RN

idad.cauchy


and if

R RN

J(x)|x|2 dx < +∞ then Z ³ ´ 2 ˆ D J (0) = − ij

11

xi xj J(x) dx, RN

¢ ¢ ¡ ¡ therefore D2 Jˆ ij (0) = 0 when i 6= j and D2 Jˆ ii (0) 6= 0. Hence the Hessian matrix of Jˆ at the origin is given by ¶ µ Z 1 2 2 ˆ |x| J(x) dx · Id. D J(0) = − N RN ˆ iii) If J(ξ) = 1 − A|ξ|α + o(|ξ|)α then necessarily α ∈ (0, 2], and if J has a first momentum, then α 6= 1 . Finally, if α = 2, then ¡ ¢ A · Id = −(1/2) D2 Jˆ (0). ij

Proof. Points i) and ii) are rather straightforward (recall that J is radially symmetric). Now we turn to iii). Let us recall a well-known probability lemma that says that if Jˆ has an expansion of the form, ˆ = 1 + iha, ξi − 1 hξ, Bξi + o(|ξ|2 ), J(ξ) 2 then J has a second momentum and we have Z Z ai = xi J(x)dx, Bij = xi xj J(x)dx < ∞. Thus if iii) holds for some α > 2, it would turn out that the second moment of J is null, which imply that J ≡ 0, a contradiction. Finally, when α = 2, then clearly ¡ 2 would ¢ ˆ Bij = − D J ij (0) hence the result since by symmetry, the Hessian is diagonal. ¤ Now, we prove existence and uniqueness of solutions using the Fourier transform. 1 Theorem 3. Let u0 ∈ L11.2 (RN ) such that uˆ0 ∈ L1 (RN ). There exists a unique solution u ∈ C 0 ([0, ∞); L1 (RN )) of (2.1), and it is given by ˆ

uˆ(ξ, t) = e(J(ξ)−1)t uˆ0 (ξ). Proof. We have

Z

ut (x, t) = J ∗ u − u(x, t) =

J(x − y)u(y, t) dy − u(x, t). RN

Applying the Fourier transform to this equation we obtain ˆ − 1). uˆt (ξ, t) = uˆ(ξ, t)(J(ξ) Hence, ˆ

uˆ(ξ, t) = e(J(ξ)−1)t uˆ0 (ξ).

12

2. THE LINEAR PROBLEM ˆ

Since uˆ0 ∈ L1 (RN ) and e(J(ξ)−1)t is continuous and bounded, the result follows by taking the inverse of the Fourier transform. ¤

.Four.Class

11.2

Remark 4. One can also understand solutions of (2.1) directly in Fourier variables. This concept of solution is equivalent to the integral one in the original variables under our hypotheses on the initial condition. 11.2

Now we prove a lemma concerning the fundamental solution of (2.1).

fundamental

decomp

Lemma 5. Let J ∈ S(RN ), the space of rapidly decreasing functions. The fundamen11.2 11.2 tal solution of (2.1), that is the solution of (2.1) with initial condition u0 = δ0 , can be decomposed as (2.3)

w(x, t) = e−t δ0 (x) + v(x, t), 11.2

with v(x, t) smooth. Moreover, if u is a solution of (2.1) it can be written as Z u(x, t) = (w ∗ u0 )(x, t) = w(x − z, t)u0 (z) dz. RN

Proof. By the previous result we have ˆ − 1). wˆt (ξ, t) = w(ξ, ˆ t)(J(ξ) Hence, as the initial datum verifies uˆ0 = δˆ0 = 1, ˆ

ˆ

w(ξ, ˆ t) = e(J(ξ)−1)t = e−t + e−t (eJ(ξ)t − 1). The first part of the lemma follows applying the inverse Fourier transform in S(RN ). 11.2 To finish the proof we just observe that w ∗ u0 is a solution of (2.1) (just use Fubini’s theorem) with (w ∗ u0 )(x, 0) = u0 (x). ¤ en.L^1

.cauchy.uno

ˆ → 0 (since J ∈ L1 (RN )) Remark 6. The above proof together with the fact that J(ξ) decomp shows that if Jˆ ∈ L1 (RN ) then the same decomposition (2.3) holds and the result also applies. Next, we prove the first part of our main result. 11.2

Theorem 7. Let u be a solution of (2.1) with u0 , uˆ0 ∈ L1 (RN ). If ˆ = 1 − A|ξ|α + o(|ξ|α ), J(ξ)

ξ → 0,

the asymptotic behavior of u(x, t) is given by lim tN/α max |u(x, t) − v(x, t)| = 0,

t→+∞

x

where v is the solution of vt (x, t) = −A(−∆)α/2 v(x, t) with initial condition v(x, 0) = u0 (x).

d.1.N.alpha


13

Proof. As in the proof of the previous lemma we have ˆ − 1). uˆt (ξ, t) = uˆ(ξ, t)(J(ξ) Hence ˆ

uˆ(ξ, t) = e(J(ξ)−1)t uˆ0 (ξ). On the other hand, let v(x, t) be a solution of vt (x, t) = −A(−∆)α/2 v(x, t), with the same initial datum v(x, 0) = u0 (x). Solutions of this equation are understood in the sense that α vˆ(ξ, t) = e−A|ξ| t uˆ0 (ξ). Hence in Fourier variables, Z Z ¯³ ¯ ´ ˆ ¯ t(J(ξ)−1) ¯ −A|ξ|α t −e uˆ0 (ξ)¯ dξ |ˆ u − vˆ|(ξ, t) dξ = ¯ e RN Z RN ¯³ ¯ ´ α ˆ ¯ t(J(ξ)−1) ¯ ≤ − e−A|ξ| t uˆ0 (ξ)¯ dξ ¯ e |ξ|≥r(t) Z ¯³ ¯ ´ ˆ ¯ t(J(ξ)−1) ¯ −A|ξ|α t + −e uˆ0 (ξ)¯ dξ = I + II. ¯ e |ξ|r(t)

α

e−A|η| → 0, |η|>r(t)t1/α

as t → ∞ if we impose that (2.4)

r(t)t1/α → ∞

as t → ∞.

Now, remark that from our hypotheses on J we have that Jˆ verifies ˆ ≤ 1 − A|ξ|α + |ξ|α h(ξ), J(ξ) where h is bounded and h(ξ) → 0 as ξ → 0. Hence there exists D > 0 such that ˆ ≤ 1 − D|ξ|α , J(ξ)

for |ξ| ≤ a,

and δ > 0 such that ˆ ≤ 1 − δ, J(ξ)

for |ξ| ≥ a.

14


Therefore, I2 can be bounded by Z Z ¯ ¯ ¯ ¯ ˆ ˆ ¯ t(J(ξ)−1) ¯ ¯ t(J(ξ)−1) ¯ uˆ0 (ξ)¯ dξ ≤ uˆ0 (ξ)¯ dξ ¯e ¯e |ξ|≥r(t) Z Za≥|ξ|≥r(t) ¯ ¯ ¯ ¯ ˆ ˆ ¯ t(J(ξ)−1) ¯ ¯ t(J(ξ)−1) ¯ + uˆ0 (ξ)¯ dξ ≤ uˆ0 (ξ)¯ dξ + Ce−δt . ¯e ¯e |ξ|≥a

a≥|ξ|≥r(t)

Using this bound and changing variables, η = ξt1/α , Z ¯ −D|η|α ¯ N/α ¯e t I2 ≤ C uˆ0 (ηt−1/α )¯ dη + tN/α Ce−δt at1/α ≥|η|≥t1/α r(t) Z α ≤C e−D|η| dη + tN/α Ce−δt , |η|≥t1/α r(t)

and then tN/α I2 → 0,

cond.1.N.alpha

as t → ∞,

if (2.4) holds.

Now we estimate II as follows, Z α α ˆ N/α t |e(J(ξ)−1+A|ξ| )t − 1| e−A|ξ| t |ˆ u0 (ξ)| dξ |ξ| 0,

1 + βα + sβ < 0.

This implies that β ∈ (−1/α, −1/(α + s)) which is of course always possible. teo.cauchy.uno

As a consequence of Theorem 7, we obtain the following corollary which completes the results gathered in the main theorem. CORO

ˆ Corollary 10. If J(ξ) = 1 − A|ξ|α + o(|ξ|α ), ξ → 0, 0 < α ≤ 2, the asymptotic 11.2 behavior of solutions of (2.1) is given by C ku(·, t)kL∞ (RN ) ≤ N/α . t Moreover, the asymptotic profile is given by ¯ ¯ lim max ¯tN/α u(yt1/α , t) − ku0 kL1 GA (y)¯ = 0, t→+∞

y


teo.cauchy.uno

Proof. From Theorem 7 we obtain that the asymptotic behavior is the same as the one for solutions of the evolution given by the fractional laplacian. It is easy to check that this asymptotic behavior is exactly the one described in the statement of the corollary. Indeed, in Fourier variables we have for t → ∞ α

α

α

vˆ(t−1/α η, t) = e−A|η| uˆ0 (ηt−1/α ) −→ e−A|η| uˆ0 (0) = e−A|η| ku0 kL1 (RN ) . Therefore

¯ ¯ lim max ¯tN/α v(yt1/α , t) − ku0 kL1 GA (y)¯ = 0,

t→+∞

y

16



¤ 11.2

Now we find the decay rate in Lp of solutions of (2.1).

eo.decay.Lp

ˆ Corollary 11. Let 1 < p < ∞. If J(ξ) = 1 11.2 − A|ξ|α + o(|ξ|α ), ξ → 0, 0 < α ≤ 2, p then, the decay of the L -norm of the solution of (2.1) is given by ¡ ¢ −N 1− p1 α ku(·, t)kLp (RN ) ≤ Ct . Brezis

Proof. By interpolation, see [27], we have 1

1− 1

kukLp (RN ) ≤ kukLp 1 (RN ) kukL∞p(RN ) . 11.2

As (2.1) preserves the L1 norm, the result follows from the previous results that give the decay in L∞ of the solutions. ¤ 0.2. The Dirichlet problem. Next we consider a bounded smooth domain Ω ⊂ R and impose boundary conditions to our model. From now on we assume that J is continuous. Consider the nonlocal problem Z ut (x, t) = x ∈ Ω, t > 0, J(x − y)u(y, t) dy − u(x, t), N

111Dir12

Autov

(2.9)

RN

u(x, t) = 0,

x 6∈ Ω, t > 0,

u(x, 0) = u0 (x),

x ∈ Ω.

In this model we have that diffusion takes place in the whole RN but we impose that u vanishes outside Ω. This is the analogous of what is called Dirichlet boundary conditions for the heat como.se.toma.el.dato.Dirichlet equation. However, the boundary data is not understood in the usual sense, see Remark 17. As for the Cauchy problem we understand solutions in an integral sense, teo.L^1.Dir see Theorem 14. In this case we find an exponential decay given by the first eigenvalue of an associated problem and the asymptotic behavior of solutions is described by the unique (up to a constant) associated eigenfunction. Let λ1 = λ1 (Ω) be given by Z Z 1 J(x − y)(u(x) − u(y))2 dx dy 2 RN RN Z (2.10) λ1 = inf2 u∈L (Ω) (u(x))2 dx Ω

and φ1 an associated eigenfunction (a function where the infimum is attained).


17

111Dir12

casita2

Theorem 12. For every u0 ∈ L1 (Ω) there exists a unique solution u of (2.9) such that u ∈ C([0, ∞); L1 (Ω)). Moreover, if u0 ∈ L2 (Ω), solutions decay to zero as t → ∞ with an exponential rate

decay.2

(2.11)

ku(·, t)kL2 (Ω) ≤ ku0 kL2 (Ω) e−λ1 t .

If u0 is continuous, positive and bounded then there exist positive constants C and C ∗ such that decay.3

ku(·, t)kL∞ (Ω) ≤ C e−λ1 t

(2.12) and

conv.2

raction.Dir

¯ ¯ lim max ¯eλ1 t u(x, t) − C ∗ φ1 (x)¯ = 0.

(2.13)

t→∞

x

A solution of the Dirichlet problem is defined as follows: u ∈ C([0, ∞); L1 (Ω)) satisfying Z ut (x, t) = J(x − y)u(y, t) dy − u(x, t), x ∈ Ω, t > 0, RN

u(x, t) = 0,

x 6∈ Ω, t > 0,

u(x, 0) = u0 (x)

x ∈ Ω.

Before studying the asymptotic behavior, we shall first derive existence and uniqueness of solutions, which is a consequence of Banach’s fixed point theorem. Fix t0 > 0 and consider the Banach space © ª Xt0 = w ∈ C([0, t0 ]; L1 (Ω)) with the norm |||w||| = max kw(·, t)kL1 (Ω) . 0≤t≤t0

We will obtain the solution as a fixed point of the operator T : Xt0 → Xt0 defined by Z tZ Tw0 (w)(x, t) = w0 (x) + J (x − y) (w(y, s) − w(x, s)) dy ds, 0

RN

Tw0 (w)(x, t) = 0,

x 6∈ Ω.

Lemma 13. Let w0 , z0 ∈ L1 (Ω) and w, z ∈ Xt0 , then there exists a constant C depending on J and Ω such that |||Tw0 (w) − Tz0 (z)||| ≤ Ct0 |||w − z||| + ||w0 − z0 ||L1 (Ω) . Proof. We have Z Z |Tw0 (w)(x, t) − Tz0 (z)(x, t)| dx ≤ |w0 − z0 |(x) dx Ω Ω Z ¯Z t Z h ¯ J (x − y) (w(y, s) − z(y, s)) + ¯¯ 0 RN Ω ¯ i ¯ −(w(x, s) − z(x, s)) dy ds¯ dx.

18


Hence, taking into account that w and z vanish outside Ω, |||Tw0 (w) − Tz0 (z)||| ≤ ||w0 − z0 ||L1 (Ω) + Ct0 |||w − z|||, as we wanted to prove.

teo.L^1.Dir

¤

Theorem 14. For every u0 ∈ L1 (Ω) there exists a unique solution u, such that u ∈ C([0, ∞); L1 (Ω)). contraction.Dir

Proof. We check first that Tu0 maps Xt0 into Xt0 . Taking z0 , z ≡ 0 in Lemma 13 we get that T (w) ∈ C([0, t0 ]; L1 (Ω)). contraction.Dir

Choose t0 such that Ct0 < 1. Now taking z0 ≡ w0 ≡ u0 in Lemma 13 we get that Tu0 is a strict contraction in Xt0 and the existence and uniqueness part of the theorem follows from Banach’s fixed point theorem in the interval [0, t0 ]. To extend the solution to [0, ∞) we may take as initial data u(x, t0 ) ∈ L1 (Ω) and obtain a solution up to [0, 2t0 ]. Iterating this procedure we get a solution defined in [0, ∞). ¤ 111Dir12

Next we look for steady states of (2.9).

111Dir12

Proposition 15. u ≡ 0 is the unique stationary solution of (2.9).

onarias.Dir

111Dir12

Proof. Let u be a stationary solution of (2.9). Then Z 0= J(x − y)(u(y) − u(x)) dy, x ∈ Ω, RN R and u(x) = 0 for x 6∈ Ω. Hence, using that J = 1 we obtain that for every x ∈ RN it holds, Z u(x) = J(x − y)u(y) dy. RN

This equation, together with u(x) = 0 for x 6∈ Ω, implies that u ≡ 0.

¤

Now, let us analyze the asymptotic behavior of the solutions. As there exists a unique stationary solution, it is expected that solutions converge to zero as t → ∞. Our main concern will be the rate of convergence. Autov

First, let us look the eigenvalue given by (2.10), that is we look for the first eigenvalue of eigenvalue

eigenvalue2

(2.14)

Z u(x) −

J(x − y)u(y) dy = λ1 u(x). RN

This is equivalent to, (2.15)

Z (1 − λ1 )u(x) =

J(x − y)u(y) dy. RN

Let T : L2 (Ω) → L2 (Ω) be the operator given by Z T (u)(x) := J(x − y)u(y) dy. RN


19

In this definition we have extended by zero a function in L2 (Ω) to the whole RN . Hence we are looking for the largest eigenvalue of T . Since T is compact this eigenvalue is attained at eigenvalue some function φ1 (x) that turns out to be an eigenfunction for our original problem (2.14). Autov By taking |φ1 | instead of φ1 in (2.10) we may assume that φ1 ≥ 0 in Ω. Indeed, one simply has to use the fact that (a − b)2 ≥ (|a| − |b|)2 . eigenvalue

Next, we analyze some properties of the eigenvalue problem (2.14). eigenvpb

eigenvalue

Proposition 16. Let λ1 the first eigenvalue of (2.14) and denote by φ1 (x) a corresponding non-negative eigenfunction. Then φ1 (x) is strictly positive in Ω and λ1 is a positive simple eigenvalue with λ1 < 1. Proof. In what follows, we denote by φ¯1 the natural continuous extension of φ1 to ¯ We begin with the positivity of the eigenfunction φ1 . Assume for contradiction that Ω. the set B = {x ∈ Ω : φ1 (x) = 0} is non-void. Then, from the continuity of φ1 in Ω, we have that B is closed. We next prove that B is also open, and hence, since Ω is connected, standard topological arguments allows to concludeeigenvalue2 that Ω ≡ B yielding to a contradiction. Consider x0 ∈ B. Since φ1 ≥ 0, we obtain from (2.15) that Ω ∩ B1 (x0 ) ∈ B. Hence B is ¯ open and the result follows. Analogous arguments apply to prove that φ¯1 is positive in Ω. ¯ Assume now for contradiction that λ1 ≤ 0 and denote by M ∗ the maximum of φ¯1 in Ω ∗ ∗ and by x a point Sol.estacionarias.Dir where such maximum is attained. Assume for the moment that x ∈ Ω. From Proposition 15, one can choose x∗ in such a way that φ1 (x) 6= M ∗ in Ω ∩ B1 (x∗ ). By eigenvalue2 using (2.15) we obtain that, Z ∗ ∗ M ≤ (1 − λ1 )φ1 (x ) = J(x∗ − y)φ1 (y) < M ∗ RN ∗

and a contradiction follows. If x ∈ ∂Ω, we obtain a similar contradiction after substituting eigenvalue2 and passing to the limit in (2.15) on a sequence {xn }eigenvalue2 ∈ Ω, xn → x∗ as n → ∞. To obtain the upper bound, assume that λ1 ≥ 1. Then, from (2.15) we obtain for every x ∈ Ω that Z ∗ 0 ≥ (1 − λ1 )φ1 (x ) = J(x∗ − y)φ1 (y) RN

a contradiction with the positivity of φ1 . Finally, to prove that λ1 is a simple eigenvalue, let φ1 6= φ2 be two different eigenfunctions associated to λ1 and define ¯ C ∗ = inf{C > 0 : φ¯2 (x) ≤ C φ¯1 (x), x ∈ Ω}. The regularity of the eigenfunctions and the previous analysis shows that C ∗ is nontrivial ¯ such that φ¯2 (x∗ ) = and bounded. Moreover from its definition, there must exists x∗ ∈ Ω eigenvalue C ∗ φ¯1 (x∗ ). Define φ(x) = C ∗ φ1 (x) − φ2 (x). From the linearity of (2.14), we have that φ is ¯ ∗ ) = 0. From he positivity of the a non-negative eigenfunction associated to λ1 with φ(x eigenfunctions stated above, it must be φ ≡ 0. Therefore, φ2 (x) = C ∗ φ1 (x) and the result follows. This completes the proof. ¤

20

o.Dirichlet


Remark 17. Note that the first eigenfunction φ1 is strictly positive in Ω (with positive ¯ and vanishes outside Ω. Therefore a discontinuity occurs on continuous extension to Ω) ∂Ω and the boundary value is not taken in the usual ”classical” sense. casita2

Proof of Theorem 12. Using the symmetry of J, we have µ Z ¶ Z Z ∂ 1 2 J(x − y)[u(y, t) − u(x, t)]u(x, t) dy dx u (x, t) dx = ∂t 2 Ω RN Z RN Z 1 =− J(x − y)[u(y, t) − u(x, t)]2 dy dx. 2 RN RN Autov From the definition of λ1 , (2.10), we get Z Z ∂ 2 u (x, t) dx ≤ −2λ1 u2 (x, t) dx. ∂t Ω Ω Therefore Z Z 2 −2λ1 t u20 (x) dx u (x, t) dx ≤ e Ω

Ω decay.2

and we have obtained (2.11).

decay.3

conv.2

We now establish the decay rate and the convergence stated in (2.12) and (2.13) respectively. Consider a nontrivial and non-negative continuous initial data u0 (x) and let 11 u(x, t) be the corresponding solution to (1.1). We first note that u(x, t) is a continuous function satisfying u(x, t) > 0 for every x ∈ Ω and t > 0, and the same holds for u¯(x, t), the unique natural continuous extension of u(x, t) to Ω. This instantaneous positivity can eigenvpb be obtained by using analogous topological arguments to those in Proposition 16.

12

In order to deal with the asymptotic analysis, is more11 convenient to introduce the rescaled function v(x, t) = eλ1 t u(x, t). By substituting in (1.1), we find that the function v(x, t) satisfies Z (2.16) vt (x, t) = J(x − y)v(y, t) dy − (1 − λ1 )v(x, t). RN

12

On the other hand, we have that Cφ1 (x) is a solution of (2.16) for every C ∈ R and moreover,12it follows from the eigenfunction analysis above, that the set of stationary solutions of (2.16) is given by S∗ = {Cφ1 , C ∈ R}. Define now for every t > 0, the function C ∗ (t) = inf{C > 0 : v(x, t) ≤ Cφ1 (x), x ∈ Ω}. 12

By definition and by using the linearity of equation (2.16), we have that C ∗ (t) is a nonincreasing function. In fact, this is a consequence of the comparison principle applied to the solutions C ∗ (t1 )φ1 (x) and v(x, t) for t larger than any fixed t1 > 0. It implies that C ∗ (t1 )φ1 (x) ≥ v(x, t) for every t ≥ t1 , and therefore, C ∗ (t1 ) ≥ C ∗ (t) for every t ≥ t1 . In an analogous way, one can see that the function C∗ (t) = sup{C > 0 : v(x, t) ≥ Cφ1 (x), x ∈ Ω},


21

is non-decreasing. These properties imply that both limits exist, lim C ∗ (t) = K ∗

t→∞

and

lim C∗ (t) = K∗ ,

t→∞

and also provides the compactness of the orbits necessary in order passing to the limit (after subsequences if needed) to obtain that v(·, t + tn ) → w(·, t) as tn → ∞ uniformly on 12 compact subsets in Ω × R+ and that w(x, t) is a continuous function which satisfies (2.16). We also have for every g ∈ ω(u0 ) there holds, K∗ φ1 (x) ≤ g(x) ≤ K ∗ φ1 (x). Moreover, C ∗ (t) plays a role of a Lyapunov function and this fact allows to conclude that ω(u0 ) ⊂ S∗ and the uniqueness of the convergence profile. In more12detail, assume that g ∈ ω(u0 ) does not belong to S∗ and consider w(x, t) the solution of (2.16) with initial data g(x) and define C ∗ (w)(t) = inf{C > 0 : w(x, t) ≤ Cφ1 (x), x ∈ Ω}.

12

It is clear that W (x, t) = K ∗ φ1 (x) − w(x, t) is a non-negative continuous solution of (2.16) and it becomes strictly positive for every t > 0. This implies that there exists t∗ > 0 such that C ∗ (w)(t∗ ) < K ∗ and by the convergence, the same holds before passing to the limit. Hence, C ∗ (t∗ + tj ) < K ∗ if j is large enough and a contradiction with the properties of C ∗ (t) follows. The same arguments allow to establish the uniqueness of the convergence profile. ¤

11.Neu

0.3. The Neumann problem. Let us turn our attention to Neumann boundary conditions. We study Z ut (x, t) = J(x − y)(u(y, t) − u(x, t)) dy, x ∈ Ω, t > 0, (2.17) Ω u(x, 0) = u0 (x), x ∈ Ω. teo.L^1

Again solutions are to be understood in an integral sense, see Theorem 20. In this model we have that the integral terms take into account the diffusion inside Ω. In fact, as we have R explained the integral J(x − y)(u(y, t) − u(x, t)) dy takes into account the individuals arriving or leaving position x from other places. Since we are integrating in Ω, we are imposing that diffusion takes place only in Ω. The individuals may not enter nor leave Ω. This is the analogous of what is called homogeneous Neumann boundary conditions in the literature.

beta

Again in this case we find that the asymptotic behavior is given by an exponential decay determined by an eigenvalue problem. Let β1 be given by Z Z 1 J(x − y)(u(y) − u(x))2 dy dx 2 Ω Ω Z (2.18) β1 = infR . 2 u∈L (Ω), Ω u=0 (u(x))2 dx Ω

22


11.Neu

Concerning the asymptotic behavior of solutions of (2.17) our last result reads as follows: casita3

11.Neu

Theorem 18. For every u0 ∈ L1 (Ω) there exists a unique solution u of (2.17) such that u ∈ C([0, ∞); L1 (Ω)). This solution preserves the total mass in Ω Z Z u(y, t) dy = u0 (y) dy. Ω

R

Ω

11.Neu

1 Moreover, let ϕ = |Ω| u , then the asymptotic behavior of solutions of (2.17) is described Ω 0 2 as follows: if u0 ∈ L (Ω),

nv.exp.intr

ku(·, t) − ϕkL2 (Ω) ≤ e−β1 t ku0 − ϕkL2 (Ω) ,

(2.19)

and if u0 is continuous and bounded there exist a positive constant C such that

v.exp1.intr

ku(·, t) − ϕkL∞ (Ω) ≤ Ce−β1 t .

(2.20)

Solutions of the Neumann problem are functions u ∈ C([0, ∞); L1 (Ω)) which satisfy Z ut (x, t) = J(x − y)(u(y, t) − u(x, t)) dy, x ∈ Ω, t > 0, Ω

u(x, 0) = u0 (x)

x ∈ Ω. CERW

As in the previous chapter, see also [42], existence and uniqueness will be a consequence of Banach’s fixed point theorem. The main arguments are basically the same but we repeat them here to make this chapter self-contained. Fix t0 > 0 and consider the Banach space Xt0 = C([0, t0 ]; L1 (Ω)) with the norm |||w||| = max kw(·, t)kL1 (Ω) . 0≤t≤t0

T.2

action.lema

We will obtain the solution as a fixed point of the operator T : Xt0 → Xt0 defined by Z tZ (2.21) Tw0 (w)(x, t) = w0 (x) + J (x − y) (w(y, s) − w(x, s)) dy ds. 0

Ω

The following lemma is the main ingredient in the proof of existence. Lemma 19. Let w0 , z0 ∈ L1 (Ω) and w, z ∈ Xt0 , then there exists a constant C depending only on Ω and J such that |||Tw0 (w) − Tz0 (z)||| ≤ Ct0 |||w − z||| + kw0 − z0 kL1 (Ω) .


23

Proof. We have Z Z |Tw0 (w)(x, t) − Tz0 (z)(x, t)| dx ≤ |w0 − z0 |(x) dx Ω Ω ¯ Z ¯Z t Z h i ¯ ¯ ¯ + ¯ J (x − y) (w(y, s) − z(y, s)) − (w(x, s) − z(x, s)) dy ds¯¯ dx. Ω

Hence

0

Ω

Z Ω

|Tw0 (w)(x, t) − Tz0 (z)(x, t)| dx ≤ kw0 − z0 kL1 (Ω) Z tZ Z tZ + |(w(y, s) − z(y, s))| dy + |(w(x, s) − z(x, s))| dx. 0

Ω

0

Ω

Therefore, we obtain, |||Tw0 (w) − Tz0 (z)||| ≤ Ct0 |||w − z||| + kw0 − z0 kL1 (Ω) , as we wanted to prove. teo.L^1

masa

¤ 11.Neu

Theorem 20. For every u0 ∈ L1 (Ω) there exists a unique solution u of (2.17) such that u ∈ C([0, ∞); L1 (Ω)). Moreover, the total mass in Ω verifies, Z Z (2.22) u(y, t) dy = u0 (y) dy. Ω

Ω

T.2

Proof. We check first that Tu0 maps Xt0 into Xt0 . From (2.21) we see that for 0 < t1 < t2 ≤ t0 , Z t2 Z kTu0 (w)(t2 ) − Tu0 (w)(t1 )kL1 (Ω) ≤ 2 |w(y, s)| dx dy ds. t1

T.2

Ω

On the other hand, again from (2.21) kTu0 (w)(t) − w0 kL1 (Ω) ≤ Ct|||w|||. These two estimates give that Tu0 (w) ∈ C([0, t0 ]; L1 (Ω)). Hence Tu0 maps Xt0 into Xt0 .

contraction.lema

Choose t0 such that Ct0 < 1. Now taking z0 ≡ w0 ≡ u0 , in Lemma 19 we get that Tu0 is a strict contraction in Xt0 and the existence and uniqueness part of the theorem follows from Banach’s fixed point theorem in the interval [0, t0 ]. To extend the solution to [0, ∞) we may take as initial data u(x, t0 ) ∈ L1 (Ω) and obtain a solution up to [0, 2t0 ]. Iterating this procedure we get a solution defined in [0, ∞). masa

We finally prove that if u is the solution, then the integral in Ω of u satisfies (2.22). Since Z tZ u(x, t) − u0 (x) = J (x − y) (u(y, s) − u(x, s)) dy ds. 0

Ω

mann.esta.2

onarias.Neu

.c.Poincare

24


We can integrate in x and apply Fubini’s theorem to obtain Z Z u(x, t) dx − u0 (x) dx = 0 Ω

Ω

and the theorem is proved.

¤

Now we study the asymptotic behavior as t → ∞. We start by analyzing the corresponding stationary problem so we consider the equation Z (2.23) 0= J(x − y)(ϕ(y) − ϕ(x)) dy. Ω

Neumann.esta.2

The only solutions are constants. In fact, in particular, (2.23) implies that ϕ is a continuous function. Set K = max ϕ(x) x∈Ω

and consider the set A = {x ∈ Ω | ϕ(x) = K}. The set A is clearly closed and non empty. We claim that it is also open in Ω. Let x0 ∈ A. We have then Z Z ¡ ¢−1 ϕ(x0 ) = J(x0 − y) dy J(x0 − y)ϕ(y) dy, Ω

Ω

and ϕ(y) ≤ ϕ(x0 ) this implies ϕ(y) = ϕ(x0 ) for all y ∈ Ω ∩ B(x0 , d), and hence A is open as claimed. Consequently, as Ω is connected, A = Ω and ϕ is constant. We have proved the following proposition: 11.Neu

Proposition 21. Every stationary solution of (2.17) is constant in Ω. Now we prove the exponential rate of convergence to steady states of solutions in L2 . Let us take β1 as Z Z 1 J(x − y)(u(y) − u(x))2 dy dx 2 Ω Ω Z . (2.24) β1 = infR u∈L2 (Ω), Ω u=0 2 (u(x)) dx Ω

It is clear that β1 ≥ 0. Let us prove that β1 is in fact strictly positive. To this end we consider the subspace of L2 (Ω) given by the orthogonal to the constants, H = hctsi⊥ and the symmetric (self-adjoint) operator T : H 7→ H given by Z Z T (u) = J(x − y)(u(x) − u(y)) dy = − J(x − y)u(y) dy + A(x)u(x). Ω

Ω

Note that T is the sum of an inversible operator and a compact operator. Since T is symmetric we have that its spectrum verifies σ(T ) ⊂ [m, M ], where m=

inf

hT u, ui

u∈H, kukL2 (Ω) =1

ate.on.beta

ate.on.beta


25

and M=

sup

hT u, ui,


Brezis

see [27]. Remark that m = =

inf

hT u, ui Z Z J(x − y)(u(x) − u(y)) dy u(x) dx


inf


Ω

Ω

= β1 . Then m ≥ 0. Now we just observe that m > 0. Brezis

In fact, if not, as m ∈ σ(T ) (see [27]), we have that T : H 7→ H is not inversible. Using Fredholm’s alternative this implies that there exists a nontrivial u ∈ H such that T (u) = 0, but then u must be constant in Ω. This is a contradiction with the fact that H is orthogonal to the constants. To study the asymptotic behavior of the solutions we need an upper estimate on β1 . mejor.c.Poincare

Lemma 22. Let β1 be given by (2.24) then Z (2.25)

J(x − y) dy.

β1 ≤ min x∈Ω

Proof. Let

Ω

Z A(x) =

J(x − y) dy. Ω

Since Ω is compact and A is continuous there exists a point x0 ∈ Ω such that A(x0 ) = min A(x). x∈Ω

For every ε small let us choose two disjoint balls of radius ε contained in Ω, B(x1 , ε) and B(x2 , ε) in such a way that xi → x0 as ε → 0. We use uε (x) = χB(x1 ,ε) (x) − χB(x2 ,ε) (x)

26


mejor.c.Poincare

as a test function in the definition of β1 , (2.24). Then we get that for every ε small it holds Z Z 1 J(x − y)(uε (y) − uε (x))2 dy dx 2 Ω Ω Z β1 ≤ (uε (x))2 dx Ω Z Z Z 2 J(x − y)uε (y) uε (x) dy dx A(x)uε (x) dx − Ω Ω Ω Z = (uε (x))2 dx Z ZΩ Z 2 A(x)uε (x) dx − J(x − y)uε (y) uε (x) dy dx Ω Ω Ω = . 2|B(0, ε)| Using the continuity of A and the explicit form of uε we obtain Z A(x)u2ε (x) dx lim Ω = A(x0 ) ε→0 2|B(0, ε)| and

Z Z J(x − y)uε (y) uε (x) dy dx lim

Ω

Ω

2|B(0, ε)|

ε→0

estimate.on.beta

= 0.

Therefore, (2.25) follows.

¤

Now let us prove the exponential convergence of u(x, t) to the mean value of the initial datum.

al.conv.Exp conv.exp

11.Neu

Theorem 23. For every u0 ∈ L2 (Ω) the solution u(x, t) of (2.17) satisfies (2.26)

ku(·, t) − ϕkL2 (Ω) ≤ e−β1 t ku0 − ϕkL2 (Ω) .

Moreover, if u0 is continuous and bounded, there exists a positive constant C > 0 such that, conv.exp1

(2.27)

ku(·, t) − ϕkL∞ (Ω) ≤ Ce−β1 t .

mejor.c.Poincare

Here β1 is given by (2.24). Proof. Let

Z 1 H(t) = (u(x, t) − ϕ)2 dx. 2 Ω mejor.c.Poincare Differentiating with respect to t and using (2.24) and the conservation of the total mass, we obtain Z Z Z 1 2 0 J(x − y)(u(y, t) − u(x, t)) dy dx ≤ −β1 (u(x, t) − ϕ)2 dx. H (t) = − 2 Ω Ω Ω


Hence

27

H 0 (t) ≤ −2β1 H(t).

Therefore, integrating we obtain, w.est

H(t) ≤ e−2β1 t H(0),

(2.28)

conv.exp

and (2.26) follows.

conv.exp1

In order to prove (2.27) let w(x, t) denote the difference w(x, t) = u(x, t) − ϕ. We seek for an exponential estimate in L∞ of11.Neu the decay of w(x, t). The linearity of the equation implies that w(x, t) is a solution of (2.17) and satisfies Z t Z −A(x)t −A(x)t A(x)s w(x, t) = e w0 (x) + e e J(x − y)w(y, s) dy ds. 0 Ω R conv.exp Recall that A(x) = Ω J(x − y)dx. By using (2.26) and the Holder inequality it follows that Z t −A(x)t −A(x)t |w(x, t)| ≤ e w0 (x) + Ce eA(x)s−β1 s ds. 0

Integrating this inequality, we obtain that the solution lema.estimate.on.beta w(x, t) decays to zero exponentially conv.exp1 fast and moreover, it implies (2.27) thanks to Lemma 22. ¤

CHAPTER 3

Refined asymptotics

ear.refined

localnliviu

ignat

In this chapter we find refined asymptotic expansions for solutions to the nonlocal evolution equation ( ut (x, t) = J ∗ u − u(x, t), t > 0, x ∈ Rd , (3.1) u(x, 0) = u0 (x), x ∈ Rd , R where J : Rd → R verifies Rd J(x)dx = 1. For the heat equation a precise asymptotic expansion in terms of the fundamental soDZ lution and its derivatives was found in [52]. In fact, if Gt denotes the fundamental solution 2 of the heat equation, namely, Gt (x) = (4πt)−d/2 e−|x| /(4t) , under adequate assumptions on the initial condition, we have, ° ° ´ X (−1)|α| ³ Z ° ° α α (3.2) u0 (x)x ∂ Gt ° ≤ Ct−A °u(x, t) − α! Lq (Rd ) Rd |α|≤k

DZ

with A = ( d2 )( (k+1) + (1 − 1q )). As pointed out by the authors in [52], the same asymptotic d s expansion can be done in a more general setting, dealing with the equations ut = −(−∆) 2 u, s > 0. Now we need to introduce some notation. We will say that f ∼ g as ξ ∼ 0 if |f (ξ) − g(ξ)| = o(g(ξ)) when ξ → 0 and, to simplify, f ≤ g if there exists a constant c independent of the relevant quantities such that f ≤ cg. In the sequel we denote by Jb the Fourier transform of J. ignat

Our main objective here is to study if an expansionChChR analogous to (3.2) holds for the nonlocalnliviu non-local problem (3.1). Concerning the first term, in [34] it is proved that if J verifies nonlocalnliviu b − 1 ∼ −|ξ|s as ξ ∼ 0, then the asymptotic behavior of the solution to (3.1), J(ξ) u(x, t), is given by d lim t s max |u(x, t) − v(x, t)| = 0, t→+∞

x

s

where v is the solution of vt (x, t) = −(−∆) 2 v(x, t) with initial condition v(x, 0) = u0 (x). d As a consequence, the decay rate is given by ku(·, t)kL∞ (Rd ) ≤ C t− s and the asymptotic profile is as follows, ° ° ³Z ´ ° ° d 1 s ° s u(yt s , t) − t u G (y) lim ° 0 ° ∞ d = 0, t→+∞ ° d R

29

L (R )

30

3. REFINED ASYMPTOTICS

cs (ξ) = e−|ξ|s . where Gs (y) satisfies G

nonlocalnliviu

Here we find a complete expansion for u(x, t), a solution to (3.1), in terms of the derivatives of the regular part of the fundamental solution, Kt . As we have mentioned, the nonlocalnliviu fundamental solution w(x, t) of problem (3.1) satisfies w(x, t) = e−t δ0 (x) + Kt (x), where the function Kt (the regular part of the fundamental solution) is given by b ct (ξ) = e−t (etJ(ξ) K − 1).

ChChR

In contrast with the previous analysis done in [34] where the long time behavior is studied in the L∞ (Rd )-norm, here we also consider Lq (Rd ) norms for q ≥ 1. The cases 2 ≤ q ≤ ∞ are derived from Hausdorff-Young’s inequality and Plancherel’s identity. The other cases 1 ≤ q < 2 are more tricky. They are reduced to L2 -estimates on the momenta of ∂ α Kt and therefore more restrictive assumptions on J have to be imposed. Theorem 24. Let s and m be positive and such that

teo.1 hyp1

b − 1 ∼ −|ξ|s , J(ξ)

(3.3)

ξ∼0

and hyp2

b |J(ξ)| ≤

(3.4)

1 , |ξ|m

|ξ| → ∞.

Then for any 2 ≤ q ≤ ∞ and k + 1 < m − d there exists a constant C = C(q, k)k|x|k+1 u0 kL1 (Rd ) such that

.expans.teo

(3.5)

° ° ´ X (−1)|α| ³ Z ° ° α α u0 (x)x ∂ Kt ° ≤ Ct−A °u(x, t) − q d α! L (R ) Rd |α|≤k

for all u0 ∈ L1 (Rd , 1 + |x|k+1 ). Here A =

(k+1) s

+ ds (1 − 1q ).

Remark 25. The condition k + 1 < m − d guarantees that all the partial derivatives ∂ Kt of order |α| = k + 1 make sense. In addition if Jb decays at infinity faster than any polynomial, α

hyp3

(3.6)

c(m) b ∀ m > 0, ∃ c(m) such that |J(ξ)| ≤ , |ξ|m

|ξ| → ∞,

asymp.expans.teo

then the expansion (3.5) holds for all k. To deal with Lq -norms for 1 ≤ q < 2 we have to impose more restrictive assumptions.


31

hyp1

aso.dificil

Theorem 26. Assume that J satisfies (3.3) with [s] > d/2 and that for any m ≥ 0 and α there exists c(m, α) such that

prop2.intro

(3.7)

b |∂ α J(ξ)| ≤

c(m, α) , |ξ|m

|ξ| → ∞. asymp.expans.teo

Then for any 1 ≤ q < 2, the asymptotic expansion (3.5) holds.

b Note that, when J has an expansion of the form J(ξ)−1 ∼ −|ξ|2 as ξ ∼ 0 (this happens for example if J is compactly supported), then the decay rate in L∞ of the solutions to the nonlocalnliviu d as t− 2 ). non-local problem (3.1) and the heat equation coincide (in both cases they decay ChChR Moreover,teo.1 the first order term also coincide (in both cases it is a Gaussian). See [34] and Theorem 24. 0.4. Estimates on Kt . To prove our result we need some estimates on the kernel Kt . In this subsection we obtain the long time behavior of the kernel Kt and its derivatives. The behavior of Lq (Rd )-norms with 2 ≤ q ≤ ∞ follows by Hausdorff-Young’s inequality in the case q = ∞ and Plancherel’s identity for q = 2. However the case 1 ≤ q < 2 is more tricky. In order to evaluate the L1 (Rd )-norm of the kernel Kt we use the following inequality carlson2

(3.8)

1−

d

d

n 2n 2n kf kL1 (Rd ) ≤ kf kL2 (R d ) k|x| f kL2 (Rd ) ,

which holds for n > d/2 and which is frequently attributed to Carlson (see for instance BB [13]). The use of the above inequality with f = Kt imposes that |x|n ∂ α Kt belongs to L2 (Rd ). To guarantee that propertycase1 and to obtain the decay rate for the L2 (Rd )-norm of |x|n ∂ α Kt we will impose in Lemma 28 additional hypotheses. The following lemma gives us the decay rate of the Lq (Rd )-norms of the kernel Kt and its derivatives for 2 ≤ q ≤ ∞ . grande2

hyp1

hyp2

Lemma 27. Let 2 ≤ q ≤ ∞ and J satisfying (3.3) and (3.4). Then for all indexes α such that |α| < m − d there exists a constant c(q, α) such that d

1

k∂ α Kt kLq (Rd ) ≤ c(q, α) t− s (1− q )−

|α| s

holds for sufficiently large t. hyp3 Moreover, if J satisfies (3.6) then the same result holds with no restriction on α. grande2

Proof of Lemma 27. We consider the cases q = 2 and q = ∞. The other cases follow by interpolation. We denote by e.s. the exponentially small terms. b ct (ξ) = e−t (etJ(ξ) First, let us consider the case q = ∞. Using the definition of Kt , K −1), we get, for any x ∈ Rd , Z b α −t |ξ||α| |etJ(ξ) − 1| dξ. |∂ Kt (x)| ≤ e Rd

32


Using that |ey − 1| ≤ 2|y| for |y| small, say |y| ≤ c0 , we obtain that 2t b b |etJ(ξ) − 1| ≤ 2t|J(ξ)| ≤ m |ξ| 1

for all |ξ| ≥ h(t) = (c0 t) m . Then Z Z b −t |α| tJ(ξ) −t e |ξ| |e − 1| dξ ≤ te |ξ|≥h(t)

|ξ|≥h(t)

|ξ||α| dξ ≤ te−t c(m − |α|) m |ξ|

provided that |α| < m − d. hyp3 Is easy to see that if (3.6) holds no restriction on the indexes α has to be assumed. It remains to estimate

Z b

−t

|ξ||α| |etJ(ξ) − 1|dξ.

e −t

R

|ξ|≤h(t)

|ξ||α| dξ is exponentially small, so we concentrate on Z b −t I(t) = e |etJ(ξ) ||ξ||α| dξ.

We observe that the term e

|ξ|≤h(t)

|ξ|≤h(t)

Now, let us choose R > 0 such that cond1444

b |J(ξ)| ≤1−

(3.9)

|ξ|s for all |ξ| ≤ R. 2

Once R is fixed, there exists δ > 0 with cond2555

b |J(ξ)| ≤ 1 − δ for all |ξ| ≥ R.

(3.10) Then

Z −t

Z b tJ(ξ)

|α|

−t

b

|I(t)| ≤ e |e ||ξ| dξ + e |etJ(ξ) ||ξ||α| dξ |ξ|≤R R≤|ξ|≤h(t) Z Z b t(|J(ξ)|−1) |α| −tδ ≤ e |ξ| dξ + e |ξ||α| dξ |ξ|≤R R≤|ξ|≤h(t) Z t|ξ|s ≤ e− 2 |ξ||α| + e.s. |ξ|≤R Z |α| |η|s |α| d − s − ds − 2 = t e |η||α| + e.s. ≤ t− s − s . 1 |η|≤Rt s

Now, for q = 2, by Plancherel’s identity we have Z b α 2 −2t |etJ(ξ) − 1|2 |ξ|2|α| dξ. k∂ Kt kL2 (Rd ) ≤ e Rd


33

Putting out the exponentially small terms, it remains to estimate Z b |et(J(ξ)−1) |2 |ξ|2|α| dξ, |ξ|≤R

cond1444 where R is given by (3.9). The behavior of Jb near zero gives Z Z 2|α| d s b t(J(ξ)−1) 2 2|α| |e | |ξ| dξ ≤ e−t|ξ| |ξ|2|α| dξ ≤ t− s − s , |ξ|≤R

|ξ|≤R

which finishes the proof.

¤

Once the case 2 ≤ q ≤ ∞ has been analyzed the next step is to obtain similar decay rates for the Lq -norms with 1 ≤ q < 2. These estimates follow from an L1 -estimate and interpolation. case1

prop2.intro

Lemma 28. Assume that J verifies (3.7) and b − 1 ∼ −|ξ|s , J(ξ)

ξ∼0

with [s] > d/2. Then for any index α = (α1 , . . . , αd ) l1

k∂ α Kt kL1 (Rd ) ≤ t−

(3.11)

|α| s

.

Moreover, for 1 < q < 2 we have d

1

k∂ α Kt kLq (Rd ) ≤ t− s (1− q )−

|α| s

for large t. Remark 29. There is no restriction on s if J is such that b |∂ α J(ξ)| ≤ min{|ξ|s−|α| , 1}, |ξ| ≤ 1. b = 1 − |ξ|s in a neighborhood of the origin. This happens if s is a positive integer and J(ξ) Remark 30. The case α = (0, . . . , 0) can be easily treated when J is nonnegative. As a consequence of the mass conservation (just integrate the equation and use Fubini’s ChChR theorem, see [34]), Z w(x, t) = 1, Rd

we obtain

Z |Kt | ≤ 1 l1

Rd

and therefore (3.11) follows with α = (0, . . . , 0). prop2.intro

Remark 31. The condition (3.7) imposed on J is satisfied, for example, for any smooth, compactly supported function J.

34


case1

Proof of Lemma 28. The estimates for 1 < q < 2 follow from the cases q = 1 and grande2 q = 2 (see Lemma 27) using interpolation. carlson2

To deal with q = 1, we use inequality (3.8) with f = ∂ α Kt and n such that [s] ≥ n > d/2. We get 1−

d

d

n α 2n 2n k∂ α Kt kL1 (Rd ) ≤ k∂ α Kt kL2 (R d ) k|x| ∂ Kt kL2 (Rd ) .

The condition n ≤ [s] guarantees that ∂ξnj Jb makes sense near ξ = 0 and thus the derivatives ct , j = 1, . . . , d, exist. Observe that the moment of order n of Kt imposes the existence ∂n K ξj

ct , j = 1, . . . , d. of the partial derivatives ∂ξnj K grande2

In view of Lemma 27 we obtain d

|α| d d k∂ α Kt kL1 (Rd ) ≤ t−( 2s + s )(1− 2n ) k|x|n ∂ α Kt kL2n2 (Rd ) .

Thus it is sufficient to prove that n

d

k|x|n ∂ α Kt kL2 (Rd ) ≤ t s − 2s −

|α| s

for all sufficiently large t. Observe that by Plancherel’s theorem Z Z 2n α 2 2n α 2 |x| |∂ Kt (x)| dx ≤ c(n) (x2n 1 + · · · + xd )|∂ Kt (x)| dx Rd

= c(n)

Rd d XZ j=1

decay1

Rd

ct )|2 dξ |∂ξnj (ξ α K

where ξ α = ξ1α1 . . . ξdαd . Therefore, it remains to prove that for any j = 1, . . . , d, it holds Z ct )|2 dξ ≤ t 2ns − ds − 2|α| s , for t large. (3.12) |∂ξnj (ξ α K Rd

We analyze the case j = 1, the others follow by the same arguments. Leibnitz’s rule gives n µ ¶ X n k α1 n−k c αd α2 n αc ∂ξ1 (ξ Kt )(ξ) = ξ2 . . . ξd ∂ξ1 (ξ1 )∂ξ1 (Kt )(ξ) k k=0 and guarantees that ct )(ξ)|2 ≤ ξ22α2 . . . ξ 2αd |∂ξn1 (ξ α K d

n X

ct (ξ)|2 K |∂ξk1 (ξ1α1 )|2 |∂ξn−k 1

k=0 min{n,α1 }

≤ ξ22α2 . . . ξd2αd decay1

X

2(α1 −k)

ξ1

ct (ξ)|2 . K |∂ξn−k 1

k=0

The last inequality reduces (3.12) to the following one: Z 2(α −k) ct (ξ)|2 dξ ≤ t 2ns − ds − 2|α| s K ξ1 1 ξ22α2 . . . ξd2αd |∂ξn−k 1 Rd


35

for all 0 ≤ k ≤ min{α1 , n}. Using the elementary inequality (it follows from the convexity of the log function) 2(α1 −k) 2α2 ξ2

ξ1

decay5

. . . ξd2αd ≤ (ξ12 + · · · + ξd2 )α1 −k+α2 +···+αd = |ξ|2(|α|−k)

it remains to prove that for any r nonnegative and any m such that n−min{α1 , n} ≤ m ≤ n the following inequality is valid, Z ct |2 dξ ≤ t− ds + 2s (m−r) . |ξ|2r |∂ξm1 K (3.13) I(r, m, t) = Rd

First we analyze the case m = 0. In this case Z b I(r, 0, t) = |ξ|2r |et(J(ξ)−1) − e−t |2 dξ Rd

grande2

and in view of Lemma 27 we obtain the desired decay property. prop2.intro

Observe that under hypothesis (3.7) no restriction on r is needed. In what follow we analyze the case m ≥ 1. First we recall the following elementary identity X ∂ξm1 (eg ) = eg ai1 ,...,im (∂ξ11 g)i1 (∂ξ21 g)i2 ...(∂ξm1 g)im i1 +2i2 +...+mim =m

ct (ξ) = where ai1 ,...,im are universal constants independent of g. Tacking into account that K b et(J(ξ)−1) − e−t we obtain for any m ≥ 1 that ct (ξ) ∂ξm1 K

X

b t(J(ξ)−1)

= e

ai1 ,...,im t

i1 +···+im

i1 +2i2 +...+mim =m

m Y

b ij [∂ξj1 J(ξ)]

j=1

and hence ct (ξ)|2 |∂ξm1 K

X

b 2t|J(ξ)−1|

≤ e

2(i1 +···+im )

t

i1 +2i2 +...+mim =m

m Y

b 2ij . [∂ξj1 J(ξ)]

j=1

Using that all the partial derivatives of Jb decay, as |ξ| → ∞, faster than any polinomial in |ξ|, we obtain that Z ct (ξ)|2 dξ ≤ e−δt t2m |ξ|2r |∂ξm1 K |ξ|>R

cond1444

cond2555

where R and δ are chosen as in (3.9) and (3.10). Tacking into account that n ≤ [s] and b − 1 + |ξ|s | ≤ o(|ξ|s ) as |ξ| → 0 we obtain that |J(ξ) b |∂ξj1 J(ξ)| ≤ |ξ|s−j , j = 1, . . . , n

36


for all |ξ| ≤ R. Then for any m ≤ n and for all |ξ| ≤ R the following holds ct (ξ)|2 |∂ξm1 K

X

−t|ξ|s

≤ e

2(i1 +···+im )

t

i1 +2i2 +...+mim =m

X

−t|ξ|s

≤ e

m Y

|ξ|2(s−j)ij

j=1 Pm

t2(i1 +···+im ) |ξ|

j=1

2(s−j)ij

.

i1 +2i2 +...+mim =m

Using that for any l ≥ 0

Z d

s

l

e−t|ξ| |ξ|l dξ ≤ t− s − s , Rd

we obtain

Z

X

d

|ξ|≤R

|ξ|2r |∂ξm1 Kt (ξ)|2 dξ ≤ t− s

2r

t2p(i1 ,...,id )− s

i1 +2i2 +···+mim =m

where m

p(i1 , . . . , im ) = (i1 + · · · + im ) −

1X (s − j)ij s j=1

m

1X m = j ij = . s j=1 s This ends the proof.

¤ teo.1

teo.1.caso.dificil

Now we are ready to prove Theorems 24 and 26. teo.1

teo.1.caso.dificil DZ

Proof of Theorems 24 and 26. Following [52] we obtain that the initial condition u0 ∈ L1 (Rd , 1 + |x|k+1 ) has the following decomposition ¶ X (−1)|α| µZ X α u0 = u0 x dx Dα δ0 + D α Fα α! |α|≤k

|α|=k+1

where kFα kL1 (Rd ) ≤ ku0 kL1 (Rd , |x|k+1 ) for all multi-indexes α with |α| = k + 1. decomp

nonlocalnliviu

In view of (2.3) the solution u of (3.1) satisfies

u(x, t) = e−t u0 (x) + (Kt ∗ u0 )(x).


37

The first term being exponentially small it suffices to analyze the long time behavior grande2 case1 of Kt ∗ u0 . Using the above decomposition, Lemma 27 and Lemma 28 we get ° ° ´ X (−1)|α| ³ Z ° ° u0 (x)xα dx ∂ α Kt ° ≤ ° K t ∗ u0 − α! Lq (Rd ) |α|≤k X ≤ k∂ α Kt ∗ Fα kLq (Rd ) |α|=k+1

≤

X

k∂ α Kt kLq (Rd ) kFα kL1 (Rd )

|α|=k+1 d

1

≤ t− s (1− q ) t−

(k+1) s

ku0 kL1 (Rd , |x|k+1 ) .

This ends the proof.

¤

0.5. Asymptotics for the higher order terms. Our next aim is to studyteo.1 if the higher order terms of the asymptotic expansion that we have found in Theorem 24 have some relation with the corresponding ones for the heat equation. Our next results say that the difference between them is of lower order. Again we have to distinguish between 2 ≤ q ≤ ∞ and 1 ≤ q < 2. teo.2 hyp4

teo.1

Theorem 32. Let J as in Theorem 24 and assume in addition that there exists r > 0 such that (3.14)

b − (1 − |ξ|s ) ∼ B|ξ|s+r , J(ξ)

ξ ∼ 0,

for some real number B. Then for any 2 ≤ q ≤ ∞ and |α| ≤ m − d there exists a positive constant C = C(q, d, s, r) such that the following holds liviu777

(3.15)

d

1

k∂ α Kt − ∂ α Gst kLq (Rd ) ≤ Ct− s (1− q ) t−

|α|+r s

,

cst (ξ) = exp(−t|ξ|s ). where Gst is defined by its Fourier transform G teo.4

Theorem 33. Let J be as in the above Theorem with [s] > d/2. We also assume that all the derivatives of Jb decay at infinity faster as any polinomial: b ≤ |∂ α J(ξ)|

c(m, α) , |ξ|m

ξ → ∞. liviu777

Then for any 1 ≤ q < 2 and any multi-index α = (α1 , . . . , αd ), (3.15) holds. Note that these results do not imply that the asymptotic expansion ¶ X (−1)|α| µZ α u0 (x)x ∂ α Kt α! |α|≤k

38

3. REFINED ASYMPTOTICS s

coincides with the expansion that holds for the equation ut = −(−∆) 2 u: ¶ X (−1)|α| µZ α u0 (x)x ∂ α Gst . α! |α|≤k

They only say that the corresponding terms agree up to a better order. When J is compactly supported or rapidly decaying at infinity, then s = 2 and we obtain an expansion analogous to the one that holds for the heat equation. teo.2

Proof of Theorem 32. Case 2 ≤ q ≤ ∞. Recall that we have defined Gst by its cst = exp(−t|ξ|s ). Fourier transform G We consider the case q = ∞, the case q = 2 can be handled similarly and the rest of the cases, 2 < q < ∞, follow again by interpolation. Writing each of the two terms in Fourier variables we obtain Z ¯ s¯ b α α s k∂ Kt − ∂ Gt kL∞ (Rd ) ≤ |ξ||α| ¯e−t (etJ(ξ) − 1) − e−t|ξ| ¯dξ. Rd

Let us choose a positive R such that ¯ ¯ b − 1 + |ξ|s ¯ ≤ C|ξ|r+s , ¯J(ξ)

for |ξ| ≤ R,

cond2555

satisfying (3.10) for some δ > 0. For |ξ| ≥ R all the terms are exponentially small as t → ∞. Thus the behavior of the difference ∂ α Kt − ∂ α Gt is given by the following integral: Z ¯ b s¯ I(t) = |ξ||α| ¯et(J(ξ)−1) − e−t|ξ| ¯dξ. |ξ|≤R

In view of the elementary inequality |ey − 1| ≤ c(R)|y| for all |y| ≤ R we obtain that Z ¯ s¯ s b I(t) = |ξ||α| e−t|ξ| ¯et(J(ξ)−1+|ξ| ) − 1¯dξ Z |ξ|≤R ¯ s¯ b − 1 + |ξ|s )¯dξ ≤ |ξ||α| e−t|ξ| ¯t(J(ξ) |ξ|≤R Z s ≤t |ξ||α| e−t|ξ| |ξ|s+r dξ |ξ|≤R |α| − ds − rs − s

≤t

.

This finishes the proof.

¤ teo.4

Proof of Theorem 33. Case 1 ≤ q < 2. Using the same ideas as in the proof of case1 Lemma 28 it remains to prove that for some d/2 < n ≤ [s] the following holds d

k|x|n (∂ α Kt − ∂ α Gst )kL2 (Rd ) ≤ t− 2s +

n−(|α|+r) s

.


39

Applying Plancherel’s identity the proof of the last inequality is reduced to the proof of the following one Z ct − G cst )]|2 dξ ≤ t− 2sd + n−(|α|+r) s |∂ξnj [ξ α (K , j = 1, . . . d, Rd

provided that all the above terms make sense. This means that all the partial derivatives ct and ∂ k G cs ∂ξkj K ξj t , j = 1, . . . , d, k = 0, . . . , n have to be defined. And thus we need n ≤ [s]. We consider the case j = 1 the other cases being similar. Applying again Leibnitz’s rule we get min{n,α1 }

ct |∂ξn1 [ξ α (K

X

cst )]|2 ≤ ξ22α2 . . . ξ 2αd −G d

2(α1 −k)

ξ1

ct − G cst )|2 |∂ξn−k (K 1

k=0 min{n,α1 }

X

≤

ct − G cst )|2 . |ξ|2(|α|−k) |∂ξn−k (K 1

k=0

In the following we prove that Z 2(m−m1 −r) ct − G cst )|2 dξ ≤ t− ds + s |ξ|2m1 |∂ξm1 (K Rd

for all |α| − min{n, α1 } ≤ m1 ≤ |α| and n − min{n, α1 } ≤ m ≤ n. Using that the integral outside of a ball of radius R decay exponentially, it remains to analyze the decay of the following integral Z ct − G cst )|2 dξ |ξ|2m1 |∂ξm1 (K |ξ|≤R

teo.2 ct and Gs we obtain where R is as in the proof of Theorem 32. Using the definition of K t that m X Y b mc t(J(ξ)−1) i1 +···+im b ij ∂ξ1 Kt (ξ) = e ai1 ,...,im t [∂ξj1 J(ξ)] i1 +2i2 +...+mim =m

and

X

cst (ξ) = etps (ξ) ∂ξm1 G

j=1

ai1 ,...,im ti1 +···+im

i1 +2i2 +...+mim =m

m Y

[∂ξj1 ps (ξ)]ij

j=1

s

where ps (ξ) = −|ξ| . Then 2 ct (ξ) − ∂ m G cs ≤ I1 (ξ, t) + I2 (ξ, t) |∂ξm1 K ξ1 t (ξ)|

where b t(J(ξ)−1)

I1 (ξ, t) = |e

tps (ξ) 2

−e

|

X i1 +2i2 +...+mim =m

2(i1 +···+im )

t

m Y j=1

|∂ξj1 ps (ξ)|2ij

40


and I2 (ξ, t) = e2tps (ξ)

X

t2(i1 +···+im ) ×

i1 +2i2 +...+mim =m ¯m ¯Y ¯ b ij × ¯ [∂ξj1 J(ξ)] ¯ j=1

−

m Y j=1

¯2 ¯ j ij ¯ [∂ξ1 ps (ξ)] ¯ . ¯

First, let us analyze I1 (ξ, t). Tacking into account that |∂ξj1 ps (ξ)| ≤ |ξ|s−j for all j ≤ m ≤ [s], |ξ| ≤ R, and that ¯2 ¯ s) b b ¯ t(J(ξ)−1) tps (ξ) 2 −2t|ξ|s ¯ t(J(ξ)−1+|ξ| − 1¯ |e −e | = e ¯e ¯2 ¯ −2t|ξ|s ¯ b s ¯ ≤ e ¯t(J(ξ) − 1 + |ξ| )¯ s

≤ t2 e−2t|ξ| |ξ|2(r+s) case1

the same arguments as in the proof of Lemma 28 give us the right decay. It remains to analyze I2 (ξ, t). We make use of the following elementary inequality ¯m ¯ m m ¯Y ¯ X Y ¯ ¯ bj ¯ ≤ |b1 . . . bj−1 ||aj − bj ||aj+1 . . . am |. ¯ aj − ¯ ¯ j=1

j=1

j=1

Then by Cauchy’s inequality we also have ¯ ¯2 m m m ¯Y ¯ Y X ¯ ¯ bj ¯ ≤ b21 . . . b2j−1 (aj − bj )2 a2j+1 . . . a2m . ¯ aj − ¯ ¯ j=1

j=1

j=1

b and bj = ∂ j ps (ξ) we obtain Applying the last inequality with aj = ∂ξj1 J(ξ) ξ1 X I2 (ξ, t) ≤ e2tps (ξ) t2(i1 +···+im ) g(i1 , . . . , im , ξ) i1 +2i2 +...+mim =m

where g(i, ξ) =

j−1 m Y X

b ik |∂ξk1 ps (ξ)|2ik ([∂ξk1 J(ξ)]

−

[∂ξk1 ps (ξ)]ik )2

j=1 k=1

n Y

b 2ik [∂ξk1 J(ξ)]

k=j+1

and i = (i1 , . . . , im ). Choosing eventually a smaller R we can guarantee that for |ξ| ≤ R and k ≤ [s] the following inequalities hold: b − ∂ξk ps (ξ)| ≤ |ξ|s+r−k , |∂ξk1 J(ξ) 1 b ≤ |ξ|s−k , |∂ξk1 J(ξ)| |∂ξk1 ps (ξ)| ≤ |ξ|s−k .


41

Hence, we get b ik − [∂ k ps (ξ)]ik | ≤ |∂ k J(ξ) b − ∂ k ps (ξ)|× |[∂ξk1 J(ξ)] ξ1 ξ1 ξ1 × ≤ |ξ|

iX k −1

l=0 s+r−k

b l [∂ξk ps (ξ)]ik −l−1 [∂ξk1 J(ξ)] 1

|ξ|(ik −1)(s−k) = |ξ|r |ξ|ik (s−k) .

This yields to the following estimate on the function g(i1 , . . . , im , ξ): g(i1 , . . . , im , ξ) ≤ |ξ|2r |ξ|2 and consequently Z Z I2 (t, ξ)dξ ≤ Rd

Pm

j=1 ik (s−k)

,

s

e−2t|ξ| × d R X ×

t2(i1 +···+im ) |ξ|2r+2

Pm

j=1 ik (s−k)

dξ.

i1 +2i2 +...+mim =m

case1

Making a change of variable and using similar arguments as in the proof of Lemma 28 we obtain the desired result. ¤ 0.6. A different approach. In this final subsection we obtain the first two terms in the asymptotic expansion of the solution under less restrictive hypotheses on J. teo.6

Theorem 34. Let u0 ∈ L1 (Rd ) with ub0 ∈ L1 (Rd ) and s < l be two positive numbers such that b − (1 − |ξ|s ) ∼ B|ξ|l , ξ ∼ 0, J(ξ) for some real number B. Then for any 2 ≤ q ≤ ∞

lim1

(3.16)

d

1

lim t s (1− q )+

t→∞

l−s s

l

ku(t) − v(t) − Bt[(−∆) 2 v](t)kLq (Rd ) → 0, s

lim2

where v is the solution to vt = −(−∆) 2 v with v(x, 0) = u0 (x). Moreover ° µZ ¶° ´ ° d + l −1 ³ ° 1 1 ° (3.17) lim °t s s u(yt s , t) − v(yt s , t) − Bh(y) u0 ° ° t→∞ d R

= 0, L∞ (Rd )

s where h is given by b h(ξ) = e−|ξ| |ξ|l .

asymp.expans.teo

Let us point out that the asymptotic expansion given by (3.5) involves Kt (and its derivatives) which is not explicit. On the other hand, the two-term asymptotic expansion lim1 (3.16) involves Gst , a well known explicit kernel (v is just the convolution of Gst and u0 ). However, our ideas and lim1 methods allow us to find only two terms in the latter expansion. The case 1 ≤ q < 2 in (3.16) can be also treated, but additional hypothesis on J have to be imposed.

42


teo.6

Proof of Theorem 34. The method that we use here is just to estimate the difference l ku(t) − v(t) − Bt(−∆) 2 v(t)kLq (Rd ) using Fourier variables. As before, it is enough to consider the cases q = 2 and q = ∞. We analyze the case q = ∞, the case q = 2 follows in the same manner by applying Plancherel’s identity. By Hausdorff-Young’s inequality we get l

ku(t) − v(t) − tB(−∆) 2 v(t)kL∞ (Rd ) Z ¯ ¯ ¯ ¯ \l b(t, ξ) − vb(t, ξ) − tB (−∆) 2 v(t, ξ)¯ dξ ≤ ¯u d ZR ¯ ¯ s b ¯ ¯ t(J(ξ)−1) − e−t|ξ| (1 + tB|ξ|l )||ub0 (ξ)¯ dξ. = ¯e Rd

As before, let us choose R > 0 such that b |J(ξ)| ≤1−

|ξ|s , 2

Then there exists δ > 0 such that b |J(ξ)| ≤ 1 − δ, Hence

|ξ| ≤ R. |ξ| ≥ R.

Z b

|et(J(ξ)−1) ||ub0 (ξ)|dξ ≤ e−δt kub0 kL1 (Rd ) |ξ|≥R

and

Z

Z b t(J(ξ)−1)

1 t− l ≤|ξ|≤R

|e

s

||ub0 (ξ)|dξ ≤ kub0 kL∞ (Rd ) e−t|ξ| /2 −1 l t ≤|ξ|≤R Z 1− s d d s ≤ t− s e−|ξ| /2 dξ ≤ t− s e−t l /4 . 1 1 |ξ|≥t s − l

Also

Z

Z −t|ξ|s

1 |ξ|≥t− l

s

l

e

(1 + tB|ξ| )|ub0 (ξ)|dξ ≤ kub0 kL∞ (Rd ) Z s 1− ds − sl ≤t e−|η| |η|l dξ 1−1 |η|≥t s l Z s s 1− ds − sl −t1− l /2 e e−|η| /2 |η|l dξ. ≤t 1 1

1 |ξ|≥t− l

e−t|ξ| t|ξ|l dξ

|η|≥t s − l

Therefore, we have to analyze Z I(t) =

1 |ξ|≤t− l

b

s

|et(J(ξ)−1) − e−t|ξ| (1 + tB|ξ|l )||ub0 (ξ)| dξ.


43

b = 1 − |ξ|s + B|ξ|l + |ξ|l f (ξ) where f (ξ) → 0 as |ξ| → 0. Thus We write J(ξ) I(t) ≤ I1 (t) + I2 (t) where

Z I1 (t) =

s

1

|ξ|≤t− l

l

l

e−t|ξ| |eBt|ξ| +t|ξ| f (ξ) − (1 + Bt|ξ|l + t|ξ|l f (ξ))||ub0 (ξ)| dξ

and

Z s

I2 (t) =

e−t|ξ| t|ξ|l f (ξ)|ub0 (ξ)| dξ.

1 |ξ|≤t− l

For I1 we have

Z s

I1 (t) ≤ kub0 kL∞ (Rd ) e−t|ξ| (t|ξ|l + t|ξ|l |f (ξ)|)2 dξ −1 |ξ|≤t l Z d 2l s ≤ e−t|ξ| t2 |ξ|2l dξ ≤ t− s +2− s 1 |ξ|≤t− l

and then

d

l

l

t s + s −1 I1 (t) ≤ t1− s → 0,

It remains to prove that

d

t → ∞.

l

t s + s −1 I2 (t) → 0, t → ∞. Making a change of variable we obtain Z d l −|ξ|s l − 1s −1+ s I (t) ≤ ku e |ξ| f (ξt ts b0 kL∞ (Rd ) ) dξ. 2 1 1 |ξ|≤t s − l

The integrand is dominated by kf kL∞ (Rd ) |ξ|l exp(−|ξ|s ), which belongs to L1 (Rd ). Hence, 1 as f (ξ/t s ) → 0 when t → ∞, this shows that d

lim1

and finishes the proof of (3.16). lim1

l

t s + s −1 I2 (t) → 0, lim2

Thanks to (3.16), the proof of (3.17) is reduced to show that ° µZ ¶° ° d+ l ° l 1 ° s s 2 s lim °t [(−∆) v](yt , t) − h(y) u0 ° ° ∞ t→∞ d R

= 0.

L (Rd )

For any y ∈ Rd by making a change of variables we obtain Z l 1 d l 1 s + e−|ξ| |ξ|l eiyξ ub0 (ξ/t s ). I(y, t) = t s s [(−∆) 2 v](yt s , t) = Rd

Thus, using the dominated convergence theorem we obtain ° ° Z Z ° ° 1 s ° °I(y, t) − h(y) ≤ e−|ξ| |ξ|l |ub0 (ξ/t s ) − ub0 (0)| dξ → 0 u0 ° ° d d ∞ d R

as t → ∞.

L (R )

R

¤

CHAPTER 4

Higher order problems

near.higher

Our main concern in this chapter is the study of the asymptotic behavior of solutions of a nonlocal diffusion operator of higher order in the whole RN , N ≥ 1.

nonloceq

Let us consider the following nonlocal evolution problem:  ut (x, t) = (−1)n−1 (J − 1)n (u(x, t))   Ã ∗n Id !  µ  X n¶ = (−1)n−1 (−1)n−k (J∗)k (u) (x, t), (4.1)  k  k=0   u(x, 0) = u0 (x), for x ∈ RN and t > 0.

nonloceq

Note that in our problem (4.1) we just have the iteration k-times of the nonlocal operator J ∗ u − u as right hand side of the equation. This can be seen as a nonlocal generalization of higher order equations of the form ec.local

fouriercond

vt (x, t) = −An (−∆)

(4.2)

αn 2

v(x, t),

with A and α are positive constants specified later in this section. Note that when α = 2 ec.local (4.2) is just vt (x, t) = −An (−∆)n v(x, t). Nonlocal higher order problems have been, for instance, proposed as models for periodic phase separation. Here the nonlocal character of the problem is associated with long-range interactions Hof ”particles” in the system. An NO OK example is the nonlocal Cahn-Hilliard equation (cf. e.g. [64], [73], [74]). nonloceq

Here we propose (4.1) as a model for higher order nonlocal evolution. For this model, we first prove existence and uniqueness of a solution, but our main aim is to study the nonloceq asymptotic behaviour as t → ∞ of solutions to (4.1). For a function f we denote by fˆ the Fourier transform of f and by fˇ the inverse Fourier transform of f . Our hypotheses on the convolution kernel J that we will assume throughout this chapter are: The kernel J ∈ C(RN , R) is a nonnegative, radial function with total mass equals one, J(x) dx = 1. This means that J is a radial density probability which implies that its RN ˆ ˆ = 1. Moreover, we assume that Fourier transform verifies |J(ξ)| ≤ 1 with J(0)

R

(4.3)

ˆ = 1 − A |ξ|α + o(|ξ|α ) J(ξ)

for some A > 0 and α > 0. 45

for ξ → 0,

c.existence

eo.exis.uni

46

4. HIGHER ORDER PROBLEMS

Under these conditions on J we have the following results. 0.7. Existence and uniqueness. First, we show existence and uniqueness of a solution 1 Theorem 35. Let u0 ∈ L (RN ) such that uˆ0 ∈ L1 (RN ). There exists a unique solution nonloceq u ∈ C 0 ([0, ∞); L1 (R)N ) of (4.1) that, in Fourier variables, is given by the explicit formula, n−1 (J(ξ)−1) nt ˆ

uˆ(ξ, t) = e(−1) Proof. We have ut (x, t) = (−1)n−1

Ã n µ ¶ X n k

k=0

uˆ0 (ξ). !

(−1)n−k (J∗)k (u) (x, t).

Applying the Fourier transform to this equation we obtain n µ ¶ X n n−1 k ˆ uˆt (ξ, t) = (−1) (−1)n−k (J(ξ)) uˆ(ξ, t) k k=0 ˆ − 1)n uˆ(ξ, t). = (−1)n−1 (J(ξ) Hence

n−1 (J(ξ)−1) nt ˆ

uˆ(ξ, t) = e(−1)

uˆ0 (ξ).

nt ˆ (−1)n−1 (J(ξ)−1)

Since uˆ0 (ξ) ∈ L1 (RN ) and e is continuous and bounded, uˆ(·, t) ∈ L1 (RN ) and the result follows by taking the inverse Fourier transform. ¤ nonloceq

Now we prove a lemma concerning the fundamental solution of (4.1). nonloceq

ental.carol

Lemma 36. The fundamental solution w of (4.1), that is the solution of the equation with initial condition u0 = δ0 , can be decomposed as

ecomp.carol

(4.4)

w(x, t) = e−t δ0 (x) + v(x, t),

nonloceq

with v(x, t) smooth. Moreover, if u is a solution of (4.1) it can be written as Z u(x, t) = (w ∗ u0 )(x, t) = w(x − z, t)u0 (z) dz. RN

Proof. By the previous result we have ˆ − 1)n w(ξ, wˆt (ξ, t) = (−1)n−1 (J(ξ) ˆ t). Hence, as the initial datum verifies wˆ0 = δˆ0 = 1, we get ³ ´ n +1]t nt ˆ ˆ [(−1)n−1 (J(ξ)−1) (−1)n−1 (J(ξ)−1) −t −t e −1 . w(ξ, ˆ t) = e =e +e The first part of the lemma follows applying the inverse Fourier transform. nonloceq

To finish the proof we just observe that w ∗ u0 is a solution of (4.1) with (w ∗ u0 )(x, 0) = u0 (x). ¤


sec.thm1

47

thm1

0.8. Asymptotic behavior. Proof of Theorem 37. Next, we deal with the asymptotic behavior as t → ∞. thm1

nonloceq

Theorem 37. Let u be a solution of (4.1) with u0 , uˆ0 ∈ L1 (RN ). Then the asymptotic behavior of u(x, t) is given by N

lim t αn max |u(x, t) − v(x, t)| = 0,

t→+∞

x

αn

where v is the solutionJfouriercond of vt (x, t) = −An (−∆) 2 v(x, t) with initial condition v(x, 0) = u0 (x) and A and α as in (4.3). Moreover, we have that there exists a constant C > 0 such that N

ku(., t)kL∞ (RN ) ≤ C t− αn and the asymptotic profile is given by ¯ N ¯ 1 ¯ ¯ lim max ¯t αn u(yt αn , t) − ku0 kL1 (RN ) GA (y)¯ = 0, t→+∞

y

αn n where GA (y) satisfies GÂ (ξ) = e−A |ξ| .

thm1

Now we prove the first part of Theorem 37. thm21

nonloceq

Theorem 38. Let u be a solution of (4.1) with u0 , uˆ0 ∈ L1 (RN ). Then, the asymptotic behavior of u(x, t) is given by N

lim t αn max |u(x, t) − v(x, t)| = 0,

t→+∞

x

where v is the solution of vt (x, t) = −An (−∆) u0 (x).

αn 2

v(x, t), with initial condition v(x, 0) =

Proof. As in the previous section, we have in Fourier variables, ˆ − 1)n uˆ(ξ, t). uˆt (ξ, t) = (−1)n−1 (J(ξ) Hence n−1 (J(ξ)−1) nt ˆ

uˆ(ξ, t) = e(−1)

uˆ0 (ξ). αn

On the other hand, let v(x, t) be a solution of vt (x, t) = −An (−∆) 2 v(x, t), with the same initial datum v(x, 0) = u0 (x). Solutions of this equation are understood in the sense that vˆ(ξ, t) = e−A

n |ξ|αn t

uˆ0 (ξ).

48


Hence in Fourier variables Z Z ¯ ¯ nt ˆ ¯ (−1)n−1 (J(ξ)−1) ¯ −An |ξ|αn t |ˆ u − vˆ| (ξ, t) dξ = −e )uˆ0 (ξ)¯ dξ ¯(e RN RN Z ¯³ ¯ ´ nt ˆ ¯ (−1)n−1 (J(ξ)−1) ¯ −An |ξ|αn t ≤ −e uˆ0 (ξ)¯ dξ ¯ e |ξ|≥r(t) Z ¯ ¯³ ´ αn nt n ˆ ¯ (−1)n−1 (J(ξ)−1) ¯ − e−A |ξ| t uˆ0 (ξ)¯ dξ + ¯ e |ξ| 0 and a constant a such that ˆ ≤ 1 − D |ξ|α , for |ξ| ≤ a. J(ξ)


49

¯ ¯ ¯ˆ ¯ Moreover, because ¯J(ξ) ¯ ≤ 1 and J is a radial function, there exists a δ > 0 such that ˆ ≤ 1 − δ, J(ξ)

for |ξ| ≥ a.

Therefore I2 can be bounded by Z ¯ ¯ nt ˆ ¯ (−1)n−1 (J(ξ)−1) ¯ I2 ≤ uˆ0 (ξ)¯ dξ ¯e a≥|ξ|≥r(t) Z ¯ ¯ nt ˆ ¯ ¯ (−1)n−1 (J(ξ)−1) uˆ0 (ξ)¯ dξ + ¯e |ξ|≥a Z αn n n ≤ kuˆ0 kL∞ (RN ) e−D |ξ| t dξ + Ce−δ t . a≥|ξ|≥r(t)

1/(αn)

Changing variables as before, η = ξt , we get Z N N −Dn |η|αn −δ n t αn e t αn I2 ≤ kuˆ0 kL∞ (RN ) e dη + Ct 1 1 at αn ≥|η|≥r(t)t αn Z αn N n n ≤ kuˆ0 kL∞ (RN ) e−D |η| dη + Ct αn e−δ t → 0, 1 |η|≥r(t)t αn

firstcond

as t → ∞ if (4.5) holds. It remains only to estimate II. We proceed as follows Z ¯ ¯ n +An |ξ|αn ] ˆ ¯ −An |ξ|αn t ¯ t[(−1)n−1 (J(ξ)−1) II = e − 1¯ |uˆ0 (ξ)| dξ. ¯e |ξ| 0 for r small. thm21

As a consequence of Theoremthm1 38, we obtain the following corollary which completes the results gathered in Theorem 37 in the Introduction. nonloceq

Corollary 40. The asymptotic behavior of solutions of (4.1) is given by C ku(., t)kL∞ (RN ) ≤ N . t αn Moreover, the asymptotic profile is given by ¯ ¯ N 1 ¯ ¯ lim max ¯t αn u(yt αn , t) − ku0 kL1 (RN ) GA (y)¯ = 0, t→+∞

y



51

thm21

Proof. From Theorem 38 we obtain that the asymptotic behavior is the same as the one for solutions of the evolution given by a power n of the fractional Laplacian. It is easy to check that the asymptotic behavior is in fact the one described in the statement of the corollary. In Fourier variables we have n |η|αn

1

limt→∞ vˆ(ηt− αn , t) = limt→∞ e−A n |η|αn

= e−A

n |η|αn

= e−A Therefore

1

uˆ0 (ηt− αn )

uˆ0 (0) ku0 kL1 (RN ) .

¯ ¯ N 1 ¯ ¯ lim max ¯t αn v(yt αn , t) − ku0 kL1 (RN ) GA (y)¯ = 0,

t→+∞

y


¤ thm21

With similar arguments as in the proof of Theorem 38 one can prove that also the nonloceq asymptotic behavior of the derivatives of solutions u of (4.1) is the same as the one for derivatives of solutions v of the evolution ofnonloceq a power n of the fractional Laplacian, assuming sufficient regularity of the solutions u of (4.1). nonloceq

Theorem 41. Let u be a solution of (4.1) with u0 ∈ W k,1 (RN ), k ≤ αn and uˆ0 ∈ L1 (RN ). Then, the asymptotic behavior of Dk u(x, t) is given by ¯ ¯ N +k lim t αn max ¯Dk u(x, t) − Dk v(x, t)¯ = 0, t→+∞

x

where v is the solution of vt (x, t) = −An (−∆) u0 (x).

αn 2

v(x, t) with initial condition v(x, 0) =

Proof. We begin again by transforming our problem for u and v in a problem for the corresponding Fourier transforms uˆ and vˆ. For this we consider ¯ ¯ ¯ k ¯ ¯ \ k ∨ ∨¯ k \ k ¯ ¯ max D u(x, t) − D v(x, t) = max ¯(D u(ξ, t)) − (D v(ξ, t)) ¯ x ξ Z ¯ Z ¯ ¯ \ ¯ k v(ξ, t) dξ = ≤ |ξ|k |ˆ u(ξ, t) − vˆ(ξ, t)| dξ. ¯Dk u(ξ, t) − D\ ¯ RN RN R Showing RN |ξ|k |ˆ u(ξ, t) − vˆ(ξ, t)| dξ → 0 as t → ∞ works analogue to the proof of Theothm21 rem 38. The additional term |ξ|k is always dominated by the exponential terms. ¤

CHAPTER 5

A linear Neumann problem with a rescale of the kernel

.linear.neu

The purpose of this chapter is to show that the solutions of the usual Neumann boundary value problem for the heat equation can be approximated by solutions of a sequence of nonlocal “Neumann” boundary value problems.

calor.I

Neumann

Given a bounded, connected and smooth domain Ω, one of the most common boundary conditions that has been imposed in the literature to the heat equation, ut = ∆u, is the Neumann boundary condition, ∂u/∂η(x, t) = g(x, t), x ∈ ∂Ω, which leads to the following classical problem,  ut − ∆u = 0 in Ω × (0, T ),    ∂u =g on ∂Ω × (0, T ), (5.1)  ∂η   u(x, 0) = u0 (x) in Ω. Now we propose a nonlocal “Neumann” boundary value problem, namely Z Z ¡ ¢ (5.2) ut (x, t) = J(x − y) u(y, t) − u(x, t) dy + G(x, x − y)g(y, t) dy, RN \Ω

Ω

where G(x, ξ) is smooth and compactly supported in ξ uniformly in x. Recall that in the previous chapter we have considered homogeneous boundary data, that is, g ≡ 0. In this model the first integral takes into account the diffusion inside Ω. In fact, as we R have explained, the integral J(x−y)(u(y, t)−u(x, t)) dy takes into account the individuals arriving or leaving position x from or to other places. Since we are integrating in Ω, we are imposing that diffusion takes place only in Ω. The last term takes into account the prescribed flux of individuals that enter or leave the domain. Neumann

The nonlocal Neumann model (5.2) and the Neumann problem for the heat equation calor.I (5.1) share many properties. For example, a comparison principle holds for both equations when G is nonnegative and the asymptotic behavior of their solutions as t → ∞ is similar, CERW see [42]. Neumann

Existence and uniqueness of solutions of (5.2) with general G is proved by a fixed point argument. Also, we prove a comparison principle when G ≥ 0. 53

54

5. A LINEAR NEUMANN PROBLEM WITH A RESCALE OF THE KERNEL

0.9. Existence and uniqueness. In this chapter we deal with existence and uniqueNeumann ness of solutions of (5.2). Our result is valid in a general L1 setting.

mann.Lineal

Theorem 42. Let Ω be a bounded domain. Let J ∈ L1 (RN ) and G ∈ L∞ (Ω × RN ). 1 N For every u0 ∈ L1 (Ω) and g ∈ L∞ loc ([0, ∞); L (R \ Ω)) there exists a unique solution u of Neumann (5.2) such that u ∈ C([0, ∞); L1 (Ω)) and u(x, 0) = u0 (x). CER

CERW

As in [40] and [42], existence and uniqueness will be a consequence of Banach’s fixed point theorem. We follow closely the ideas of those works in our proof, so we will only outline the main arguments. Fix t0 > 0 and consider the Banach space Xt0 = C([0, t0 ]; L1 (Ω)) with the norm |||w||| = max kw(·, t)kL1 (Ω) . 0≤t≤t0

T

We will obtain the solution as a fixed point of the operator Tu0 ,g : Xt0 → Xt0 defined by Z tZ Tu0 ,g (w)(x, t) = u0 (x) + J (x − y) (w(y, s) − w(x, s)) dy ds 0 Ω (5.3) Z tZ + G(x, x − y)g(y, t) dy ds. 0

RN \Ω

The following lemma is the main ingredient in the proof of existence.

contraction

ec.contrac

teo.Neumann.Lineal

Lemma 43. Let J and G as in Theorem 42. Let g, h ∈ L∞ ((0, t0 ); L1 (RN \ Ω)) and u0 , v0 ∈ L1 (Ω). There exists a constant C depending only on Ω, J and G such that for w, z ∈ Xt0 , ¡ ¢ (5.4) |||Tu0 ,g (w) − Tv0 ,h (z)||| ≤ ku0 − v0 kL1 + Ct0 |||w − z||| + kg − hkL∞ ((0,t0 );L1 (RN \Ω)) . Proof. We have Z Z |Tu0 ,g (w)(x, t) − Tv0 ,h (z)(x, t)| dx ≤ |u0 (x) − v0 (x)| dx Ω Ω ¯ Z ¯Z t Z h i ¯ ¯ ¯ + ¯ J (x − y) (w(y, s) − z(y, s)) − (w(x, s) − z(x, s)) dy ds¯¯ dx Ω

0

Ω

Z Z tZ +

|G(x, x − y)||g(y, s) − h(y, s)| dy ds dx. Ω

0

RN \Ω

ec.contrac

Therefore, we obtain (5.4).

¤

teo.Neumann.Lineal

Proof of Theorem 42. Let T = Tu0 ,g . We check first that T maps Xt0 into Xt0 . T From (5.3) we see that for 0 ≤ t1 < t2 ≤ t0 , Z t2 Z Z t2 Z kT (w)(t2 ) − T (w)(t1 )kL1 (Ω) ≤ A |w(y, s)| dy ds + B |g(y, s)| dy ds. t1

Ω

t1

RN \Ω


55

T

On the other hand, again from (5.3)

© ª kT (w)(t) − u0 kL1 (Ω) ≤ Ct |||w||| + kgkL∞ ((0,t0 );L1 (RN \Ω)) .

These two estimates give that T (w) ∈ C([0, t0 ]; L1 (Ω)). Hence T maps Xt0 into Xt0 . contraction Choose t0 such that Ct0 < 1. From Lemma 43 we get that T is a strict contraction in Xt0 and the existence and uniqueness part of the theorem follows from Banach’s fixed point theorem in the interval [0, t0 ]. To extend the solution to [0, ∞) we may take as initial datum u(x, t0 ) ∈ L1 (Ω) and obtain a solution in [0, 2 t0 ]. Iterating this procedure we get a solution defined in [0, ∞). ¤ Neumann

Our next aim is to prove a comparison principle for (5.2) when J, G ≥ 0. To this end we define what we understand by sub and supersolutions. Neumann

Definition 44. A function u ∈ C([0, T ); L1 ((Ω)) is a supersolution of (5.2) if u(x, 0) ≥ u0 (x) and Z Z ¡ ¢ G(x, x − y)g(y, t) dy. J(x − y) u(y, t) − u(x, t) dy + ut (x, t) ≥ RN \Ω

Ω

Subsolutions are defined analogously by reversing the inequalities. compar.0

coro.compar

Lemma 45. Let J, G ≥ 0, u0 ≥ 0 and g ≥ 0. If u ∈ C(Ω × [0, T ]) is a supersolution to (5.2), then u ≥ 0. Neumann

Proof. Assume that u(x, t) is negative somewhere. Let v(x, t) = u(x, t) + εt with ε so small such that v is still negative somewhere. Then, if we take (x0 , t0 ) a point where v attains its negative minimum, there holds that t0 > 0 and Z vt (x0 , t0 ) = ut (x0 , t0 ) + ε > J(x − y)(u(y, t0 ) − u(x0 , t0 )) dy Ω Z = J(x − y)(v(y, t0 ) − v(x0 , t0 )) dy ≥ 0 Ω

which is a contradiction. Thus, u ≥ 0.

¤

Corollary 46. Let J, G ≥ 0 and bounded. Let u0 and v0 in L1 (Ω) with u ≥ v0 Neumann 0 ∞ 1 N and g, h ∈ L ((0, T ); L (R \ Ω)) with g ≥ h. Neumann Let u be a solution of (5.2) with initial condition u0 and flux g and v be a solution of (5.2) with initial condition v0 and flux h. Then, u≥v a.e. Proof. Let w = u − v. Then, w is a supersolution with initial datum u0 − v0 ≥ 0 and boundary datum g − h ≥ 0. contraction Using the continuity of solutions with respect to the initial J ∈ L∞ (RN ), G ∈ L∞ (Ω × RN ) we may and Neumann data (Lemma 43) and the fact that compar.0 assume that u, v ∈ C(Ω × [0, T ]). By Lemma 45 we obtain that w = u − v ≥ 0. So the corollary is proved. ¤

56


Corollary 47. Let J, G ≥ 0 and bounded. Let u ∈ C(Ω × [0, T ]) (resp. v) be a Neumann supersolution (resp. subsolution) of (5.2). Then, u ≥ v. Proof. It follows the lines of the proof of the previous corollary.

¤

0.10. Rescaling the kernel. Our main goal now is to show that the Neumann probcalor.I lem for theNeumann heat equation (5.1), can be approximated by suitable nonlocal Neumann problems like (5.2) when we rescale them appropriately.

rescales

Neumann.C.1

More precisely, for given J and G we consider the rescaled kernels µ ¶ µ ¶ 1 1 ξ ξ (5.5) Jε (ξ) = C1 N J , Gε (x, ξ) = C1 N G x, ε ε ε ε with C1−1

1 = 2

Z B(0,d)

2 J(z)zN dz,

which is a normalizing constant in order to obtain the Laplacian in the limit instead of a multiple of it. Then, we consider the solution uε (x, t) to  Z 1  ε  ut (x, t) = 2 Jε (x − y)(uε (y, t) − uε (x, t)) dy    ε Ω Z 1 (5.6) + Gε (x, x − y)g(y, t) dy,   ε N \Ω  R   uε (x, 0) = u (x). 0

We will show that uε → u in different topologies according to two different choices of the kernel G. Let us give an heuristic idea in one space dimension, with Ω = (0, 1), of why the scaling rescales involved in (5.5) is the correct one. We assume that Z ∞ Z 0 Z 1 G(1, 1 − y) dy = − G(0, −y) dy = J(y) y dy 1

−∞

0

Neumann.C.1

and, as stated above, G(x, ·) has compact support independent of x. In this case (5.6) reads Z Z ¡ ¢ 1 1 1 0 ut (x, t) = 2 Jε (x − y) u(y, t) − u(x, t) dy + Gε (x, x − y) g(y, t) dy ε 0 ε −∞ Z 1 +∞ + Gε (x, x − y) g(y, t) dy := Aε u(x, t). ε 1


57

If x ∈ (0, 1) a Taylor expansion gives that for any fixed smooth u and ε small enough, Neumann.C.1 the right hand side Aε u in (5.6) becomes 1 Aε u(x) = 2 ε

Z

1

Jε (x − y) (u(y) − u(x)) dy ≈ uxx (x) 0

and if x = 0 and ε small, Z Z 1 1 1 0 C2 Aε u(0) = 2 Jε (−y) (u(y) − u(0)) dy + Gε (0, −y) g(y) dy ≈ (ux (0) − g(0)). ε 0 ε −∞ ε Analogously, Aε u(1) ≈ (C2 /ε)(−ux (1) + g(1)). However, the proofs of our results are much more involved than simple Taylor expansions due to the fact that for each ε > 0 there are points x ∈ Ω for which the ball in which integration takes place, B(x, dε), is not contained in Ω. Moreover, when working in several space dimensions, one has to take into account the geometry of the domain.

0.11. Uniform convergence in the case g ≡ 0. Our first result deals with homogeneous boundary conditions, this is, g ≡ 0. zero

Theorem 48. Assume g ≡ 0. Let Ω be a bounded C 2+α domain for some 0 < α 0 and constants fi (¯ x) such that ( B2dε (¯ x) ∩

yN − x¯N + (

B2dε (¯ x) ∩

¡

N −1 X

¢ fi (¯ x)(yi − xi )2 > κε2+α

)

i=1

¡

yN − x¯N +

N −1 X i=1

¢ fi (¯ x)(yi − xi )2 < −κε2+α

⊂ RN \ Ω, ) ⊂ Ω.

60


Therefore Z −1 ³N X (yi − xi ) ´ x, t) Jε (x − y) θxi (¯ dy ε RN \Ω i=1 Z −1 ³N X (yi − xi ) ´ Jε (x − y) ¯ θ (¯ x , t) = dy ¡ ¢¯ xi P −1 ε 2 ¯≤κε2+α (RN \Ω)∩ ¯yN − x ¯N + N f (¯ x )(y −x ) i i i i=1 i=1

Z +

¡ ¢ P −1 2 >κε2+α yN − x ¯N + N f (¯ x )(y −x ) i i i i=1

−1 ³N X (yi − xi ) ´ dy Jε (x − y) θxi (¯ x, t) ε i=1

= I1 + I2 . If we take z = (y − x)/ε as a new variable, recalling that x¯N − xN = εs, we obtain Z N −1 X 1+α J(z)|zi | dz ≤ C κ ε |I1 | ≤ C1 |θxi (¯ x, t)| ¯ ¡ P . ¢¯ N −1 ¯zN − s+ε i=1 fi (¯x)(zi )2 ¯≤κε1+α i=1

On the other hand, I2 = C1

N −1 X i=1

Z θxi (¯ x, t)

¡ P ¢ −1 2 >κε1+α zN − s+ε N f (¯ x )(z ) i i i=1

J(z) zi dz.

Fix 1 ≤ i ≤ N − 1. Then, since symmetric, J(z) zi is an n J is¡ radially o odd function of the ¢ PN −1 x)(zi )2 > κε1+α is symmetric in that variable zi and, since the set zN − s + ε i=1 fi (¯ variable we get I2 = 0. Collecting the previous estimates the lemma is proved. ¤ We will also need the following inequality. lomitos

tegrales.II

Lemma 50. There exist K > 0 and ε¯ > 0 such that, for ε < ε¯, Z Z (y − x) (5.8) Jε (x − y)η(¯ x) · dy ≥ K Jε (x − y) dy. ε RN \Ω RN \Ω Proof. Let us put the origin at the point x¯ and take a coordinate system such that η(¯ x) = eN . Then, x = (0, −µ) with 0 < µ < dε. Then, arguing as before, Z Z (y − x) yN + µ Jε (x − y)η(¯ x) · dy = Jε (x − y) dy ε ε RN \Ω RN \Ω Z Z yN + µ yN + µ = Jε (x − y) dy + Jε (x − y) dy ε ε {yN >κε2 } RN \Ω∩{|yN |κε2 }


61

Fix c1 small such that Z Z 1 J(z) zN dz ≥ 2c1 J(z) dz. 2 {zN >0} {0κε2 } {zN >κε+ µ } ε ÃZ ! Z (5.9) = C1 J(z) zN dz − J(z) zN dz µZ

{00}

Then,

¶

Z

{00} if ε is small enough and Z 1 J(z) zN dz. K< 4 {zN >0} Case II Assume that µ ≥ c1 ε. For y in RN \ Ω ∩ B(¯ x, dε) we have yN ≥ −κε. ε Then, Z Z yN + µ Jε (x − y) dy − K Jε (x − y) dy ε RN \Ω RN \Ω Z Z ≥ (c1 − κε) Jε (x − y) dy − K Jε (x − y) dy RN \Ω RN \Ω Z ¡ ¢ = c1 − κε − K Jε (x − y) dy ≥ 0, RN \Ω

if ε is small and K< integrales.II

This ends the proof of (5.8).

zero

We now prove Theorem 48.

c1 . 2 ¤

62


zero

Proof of Theorem 48. We will use a comparison argument. First, let us look for a supersolution. Let us pick an auxiliary function v as a solution to  vt − ∆v = h(x, t) in Ω × (0, T ),      ∂v = g1 (x, t) on ∂Ω × (0, T ),  ∂η     v(x, 0) = v1 (x) in Ω. for some smooth functions h(x, t) ≥ 1, g1 (x, t) ≥ 1 and v1 (x) ≥ 0 such that the resulting v has an extension v˜ that belongs to C 2+α,1+α/2 (RN × [0, T ]), and let M be an upper bound ¯ × [0, T ]. Then, for v in Ω Z 1 ˜ vt = Lε v + (∆v − Lε v˜) + 2 Jε (x − y)(˜ v (y, t) − v˜(x, t)) dy + h(x, t). ε RN \Ω Since ∆v = ∆˜ v in Ω, we have that v is a solution to ( vt − Lε v = H(x, t, ε) in Ω × (0, T ), v(x, 0) = v1 (x) in Ω, est1

F

where by (5.7), Lemma 49 and the fact that h ≥ 1, Z 1 ˜ ε v˜) + H(x, t, ε) =(∆˜ v−L Jε (x − y)(˜ v (y, t) − v˜(x, t)) dy + h(x, t) ε2 RN \Ω ³1 Z (y − x) Jε (x − y)η(¯ x) · ≥ g1 (¯ x, t) dy ε RN \Ω ε Z h¡ (y − x¯) ¢ X Dβ v˜ ¡ (x − x¯) ¢β i ´ β + Jε (x − y) (¯ x, t) − dy + 1 − Cεα 2 ε ε N R \Ω |β|=2 Z Z ³ g (¯ ´ 1 (y − x) 1 x, t) ≥ Jε (x − y)η(¯ x) · dy − D1 Jε (x − y) dy + ε ε 2 RN \Ω RN \Ω for some constant D1 if ε is small so that Cεα ≤ 1/2. lomitos

Now, observe that Lemma 50 implies that for every constant C0 > 0 there exists ε0 such that, Z Z 1 (y − x) Jε (x − y)η(¯ x) · dy − C0 Jε (x − y) dy ≥ 0, ε RN \Ω ε RN \Ω if ε < ε0 .


est1

63

F

Now, since g = 0, by (5.7) and Lemma 49 we obtain Z h¡ (y − x¯) ¢ X Dβ u˜ ¡ (x − x¯) ¢β i β α |Fε | ≤ Cε + Jε (x − y) (¯ x, t) − dy 2 ε ε RN \Ω |β|=2 Z ≤ Cεα + C2 Jε (x − y) dy. RN \Ω

Given δ > 0, let vδ = δv. Then vδ verifies ( (vδ )t − Lε vδ = δH(x, t, ε) vδ (x, 0) = δv1 (x)

in Ω × (0, T ), in Ω.

By our previous estimates, there exists ε0 = ε0 (δ) such that for ε ≤ ε0 , |Fε | ≤ δH(x, t, ε). So, by the comparison principle for any ε ≤ ε0 it holds that −M δ ≤ −vδ ≤ wε ≤ vδ ≤ M δ. Therefore, for every δ > 0, −M δ ≤ lim inf wε ≤ lim sup wε ≤ M δ. ε→0

ε→0

and the theorem is proved.

¤

0.12. The non-homogeneous case. g 6= 0. Now we will make explicit the functions G we will deal with in the case g 6= 0. To define the first one let us introduce some notation. As before, let Ω be a bounded C domain. For x ∈ Ωε := {x ∈ Ω | dist(x, ∂Ω) < dε} and ε small enough we write x = x¯ − s d η(¯ x) where x¯ is the orthogonal projection of x on ∂Ω, 0 < s < ε and η(¯ x) is the unit exterior normal to Ω at x¯. Under these assumptions we define 2+α

italiano

(5.10)

G1 (x, ξ) = −J(ξ) η(¯ x) · ξ

for x ∈ Ωε .

Neumann.C.1

Notice that the last integral in (5.6) only involves points x ∈ Ωε since when y 6∈ Ω, x − y ∈ supp Jε implies that x ∈ Ωε . Hence the above definition makes sense for ε small. For this choice of the kernel, G = G1 , we have the following result. conv

Theorem 51. Let Ω be a bounded C 2+αcalor.I domain, g ∈ C 1+α,(1+α)/2 ((RN \ Ω) × [0, T ]), 2+α,1+α/2 u∈C (Ω × [0, T ]) the solution to (5.1), for some 0 < α < 1. Let J as before and italiano Neumann.C.1 ε G(x, ξ) = G1 (x, ξ), where G1 is defined by (5.10). Let u be the solution to (5.6). Then, sup kuε (·, t) − u(·, t)kL1 (Ω) → 0

t∈[0,T ]

as ε → 0.

64


Observe that G1 may fail to be nonnegative and hence a comparison principle may not hold. However, in this case our proof of convergence to the solution of the heat equation Neumann does not rely on comparison arguments for (5.2). If we want a nonnegative kernel G, in order to have a comparison principle, we can modify (G1 )ε by taking ¡ ¢ ˜ 1 )ε (x, ξ) = (G1 )ε (x, ξ) + κεJε (ξ) = 1 Jε (ξ) −η(¯ (G x) · ξ + κε2 ε instead. ˜ 1 )ε (x, x−y) = 1 Jε (x−y) (−η(¯ x) · (x − y) + κε2 ) Note that for x ∈ Ω and y ∈ RN \Ω, (G ε is nonnegative for ε small if we choose the constant κ as a bound for the curvature of ∂Ω, suspiro.lima conv since |x − y| ≤ d ε. As will be seen in Remark 53, Theorem 51 remains valid with (G1 )ε ˜ 1 )ε . replaced by (G 0.13. Convergence in L1 in the case G = G1 . Using the previous notations, first we prove that Fε goes to zero as ε goes to zero. fep

Lemma 52. If G = G1 then Fε (x, t) → 0

in

¡ ¢ L∞ [0, T ]; L1 (Ω)

as ε → 0. est1

F

Proof. As G = G1 = −J(ξ) η(¯ x) · ξ, for x ∈ Ωε , by (5.7) and Lemma 49, Z ¢ 1 (y − x) ¡ Fε (x, t) = Jε (x − y)η(¯ x) · g(y, t) − g(¯ x, t) dy ε RN \Ω ε Z h¡ (y − x¯) ¢ β X D u˜ ¡ (x − x¯) ¢β i β − Jε (x − y) (¯ x, t) − dy + O(εα ). 2 ε ε RN \Ω |β|=2

As g is smooth, we have that Fε is bounded in Ωε . Recalling the fact that |Ωε | = O(ε) and Fε (x, t) = O(εα ) on Ω \ Ωε we get the convergence result. ¤ conv

We are now ready to prove Theorem 51. conv

fep

Proof of Theorem 51. In the case G = G1 we have proven in Lemma 52 that Fε → 0 in L1 (Ω × [0, T ]). On the other hand, we have that wε = uε − u is a solution to wt − Lε (w) = Fε w(x, 0) = 0. ε

Let z be a solution to

zt − Lε (z) = |Fε | z(x, 0) = 0.

ε

Then −z is a solution to

zt − Lε (z) = −|Fε | z(x, 0) = 0.


65

By comparison we have that −z ε ≤ wε ≤ z ε

and z ε ≥ 0.

Integrating the equation for z ε we get Z Z Z t ε ε kz (·, t)kL1 (Ω) = z (x, t) dx = |Fε (x, s)| ds dx . Ω

fep

Ω

0

Applying Lemma 52 we get sup kz ε (·, t)kL1 (Ω) → 0

t∈[0,T ]

as ε → 0. So the theorem is proved.

uspiro.lima

¤

Remark 53. Notice that if we consider a kernel G which is a modification of G1 of the form Gε (x, ξ) = (G1 )ε (x, ξ) + A(x, ξ, ε) with Z |A(x, x − y, ε)| dy → 0 RN \Ω

conv

1

in L (Ω) as ε → 0, then the conclusion of Theorem 51 is still valid. In particular, we can take A(x, ξ, ε) = κεJε (ξ). 0.14. The case G = C J. Finally, the other “Neumann” kernel we propose is just a scalar multiple of J, that is, G(x, ξ) = G2 (x, ξ) = C2 J(ξ), where C2 is such that C2

Z

d

Z

¡ ¢ J(z) C2 − zN dz ds = 0.

(5.11) 0

{zN >s}

This choice of G is natural since we are considering a flux with a jumping probability that is a scalar multiple of the same jumping probability that moves things in the interior of the domain, J. Neumann

CERW

Several properties of solutions to (5.2) have been recently investigated in [42] in the case G = G2 for different choices of g. For the case of G2 we can still prove convergence but in a weaker sense. teo.conv.J

Theorem 54. Let Ω be a bounded C 2+αcalor.I domain, g ∈ C 1+α,(1+α)/2 ((RN \ Ω) × [0, T ]), 2+α,1+α/2 u∈C (Ω × [0, T ]) the solution to (5.1), for some 0 < α < 1. Let J as beforeNeumann.C.1 and C2 ε G(x, ξ) = G2 (x, ξ) = C2 J(ξ), where C2 is defined by (5.11). Let u be the solution to (5.6). Then, for each t ∈ [0, T ] uε (x, t) * u(x, t) as ε → 0.

∗ −weakly in L∞ (Ω)

lema.3.2.II

66


0.15. Weak convergence in L1 in the case G = G2 . First, we prove that in this case Fε goes to zero as measures. Lemma 55. If G = G2 then there exists a constant C independent of ε such that Z

T

Z |Fε (x, s)| dx ds ≤ C.

0

Ω

Moreover, Fε (x, t) * 0

as measures

as ε → 0. That is, for any continuous function θ, it holds that Z

T

Z Fε (x, t)θ(x, t) dx dt → 0

0

Ω

as ε → 0. Proof. As G F= G2 = C2 J(ξ) and g and u˜ are smooth, taking again the coordinate system of Lemma 49, we obtain Z ³ ´ 1 y N − xN Fε (x, t) = Jε (x − y) C2 g(y, t) − g(¯ x, t) ε RN \Ω ε Z N −1 X 1 (yi − xi ) − Jε (x − y) u˜xi (¯ x, t) dy ε RN \Ω ε i=1 Z X Dβ u˜(¯ x, t) h¡ (y − x¯) ¢β ¡ (x − x¯) ¢β i − Jε (x − y) − dy + O(εα ) 2 ε ε RN \Ω |β|=2 Z ³ ´ 1 y N − xN = Jε (x − y) C2 g(¯ x, t) − g(¯ x, t) ε RN \Ω ε Z N −1 X (yi − xi ) 1 Jε (x − y) u˜xi (¯ x, t) dy + O(1)χΩε + O(εα ). − ε RN \Ω ε i=1 Let

Z

³ ´ yN − xN Bε (x, t) := Jε (x − y) C2 g(¯ x, t) − g(¯ x, t) ε RN \Ω Z N −1 X (yi − xi ) u˜xi (¯ x, t) dy. − Jε (x − y) ε RN \Ω i=1


67

F

Proceeding in a similar way as in the proof of Lemma 49 we get for ε small, Z ³ ´ yN − xN Jε (x − y) C2 g(¯ x, t) − g(¯ x, t) ε RN \Ω µ ¶ Z (yN − xN ) = g(¯ x, t) Jε (x − y) C2 − dy ε (RN \Ω)∩{|yN −¯ xN |≤κε2 } µ ¶ Z (yN − xN ) + g(¯ x, t) Jε (x − y) C2 − dy ε (RN \Ω)∩{yN −¯ xN >0} µ ¶ Z (yN − xN ) − g(¯ x, t) Jε (x − y) C2 − dy ε (RN \Ω)∩{0κε2 }

(yi − xi ) dy ε

Z J(z)zi dz + O(ε)χΩε {zN −s>κε}

= I2 + O(ε)χΩε .

As in Lemma 49 we have I2 = 0. Therefore, Z ¡ ¢ Bε (x, t) = C1 g(¯ x, t) J(z) C2 − zN dz + O(ε)χΩε . {zN >s}

Now, we observe that Bε is bounded and supported in Ωε . Hence Z tZ Z Z 1 t |Fε (x, τ )| dx dτ ≤ |Bε (x, τ )| dx dτ + Ct|Ωε | + Ct|Ω|εα ≤ C. ε 0 Ω 0 Ωε This proves the first assertion of the lemma. Now, let us write for a point x ∈ Ωε x = x¯ − µη(¯ x)

with 0 < µ < dε.

For ε small and 0 < µ < dε, let dSµ be the area element of {x ∈ Ω | dist (x, ∂Ω) = µ}. Then, dSµ = dS + O(ε), where dS is the area element of ∂Ω.

ma.que.diga

68


So that, taking now µ = sε we get for any continuous test function θ, Z Z 1 T Bε (x, t)θ(¯ x, t) dx dt ε 0 Ωε Z TZ Z dZ ¡ ¢ = O(ε) + C1 g(¯ x, t)θ(¯ x, t) J(z) C2 − zN dz ds dS dt 0

∂Ω

0

{zN >s}

= O(ε) → 0 as ε → 0, since we have chosen C2 so that Z dZ 0

¡ ¢ J(z) C2 − zN dz ds = 0.

{zN >s}

Now, with all these estimates, we go back to Fε . We have 1 Fε (x, t) = Bε (x, t) + O(1)χΩε + O(εα ). ε Thus, we obtain

Z

T

Z Fε (x, t)θ(¯ x, t) dx dt → 0

0

as ε → 0.

Ωε

Now, if σ(r) is the modulus of continuity of θ, Z TZ Z TZ Fε (x, t)θ(x, t) dx dt = Fε (x, t)θ(¯ x, t) dx dt 0

Ωε T

Z

0

Z

¡ ¢ Fε (x, t) θ(x, t) − θ(¯ x, t) dx dt

+ 0 T Z

Z

Ωε

Ωε

≤

Z

T

Z

Fε (x, t)θ(¯ x, t) dx dt + Cσ(ε) 0

Ωε

|Fε (x, t)| dx dt → 0 0

as ε → 0.

Ωε

Finally, the observation that Fε = O(εα ) in Ω \ Ωε gives Z TZ Fε (x, t)θ(x, t) dx dt → 0 as ε → 0 0

Ω\Ωε

and this ends the proof.

¤

Now we prove that uε is uniformly bounded when G = G2 . Lemma 56. Let G = G2 . There exists a constant C independent of ε such that kuε kL∞ (Ω×[0,T ]) ≤ C.


69

Proof. Again we will use a comparison argument. Let us look for a supersolution. Pick an auxiliary function v as a solution to

calor.8

 vt − ∆v = h(x, t)      ∂v = g1 (x, t)  ∂η     v(x, 0) = v1 (x)

(5.12)

in Ω × (0, T ), on ∂Ω × (0, T ), in Ω.

for some smooth functions h(x, t) ≥ 1, v1 (x) ≥ u0 (x) and g1 (x, t) ≥

2 (C2 + 1) max |g(x, t)| + 1 ∂Ω×[0,T ] K

integrales.II

(K as in (5.8))

such that the resulting v has an extension v˜ that belongs to C 2+α,1+α/2 (RN × [0, T ]) and ¯ × [0, T ]. As before v is a solution to let M be an upper bound for v in Ω (

vt − Lε v = H(x, t, ε) v(x, 0) = v1 (x)

in Ω × (0, T ), in Ω,

where H verifies Z Z ³ g (¯ ´ 1 (y − x) 1 x, t) H(x, t, ε) ≥ Jε (x − y)η(¯ x) · dy − D1 Jε (x − y) dy + . ε ε 2 RN \Ω RN \Ω lomitos

So that, by Lemma 50,

¡ g1 (¯ ¢ x, t) K H(x, t, ε) ≥ − D1 ε

Z Jε (x − y) dy + RN \Ω

1 2

for ε < ε¯. Let us recall that Z C 2 ˜ ε (˜ Fε (x, t) = L u) − ∆˜ u+ Jε (x − y)g(y, t) dy ε RN \Ω Z ¡ ¢ 1 − 2 Jε (x − y) u˜(y, t) − u˜(x, t) dy. ε RN \Ω

70


F

Then, proceeding once again as in Lemma 49 we have, Z Z ¯ |g(¯ x, t)| C2 |g(¯ x, t)| (y − x) ¯¯ |Fε (x, t)| ≤ Jε (x − y) dy + Jε (x − y)¯η(¯ x) · dy ε ε ε RN \Ω RN \Ω Z α + Cε + C Jε (x − y) dy RN \Ω

h (C + 1) iZ 2 ≤ max |g(x, t)| + C Jε (x − y) dy + Cεα ∂Ω×[0,T ] ε N R \Ω Z ¢ ¡ g1 (¯ x, t) K +C Jε (x − y) dy + Cεα ≤ 2ε RN \Ω if ε < ε¯, by our choice of g1 . Therefore, for every ε small enough, we obtain |Fε (x, t)| ≤ H(x, t, ε), and, by a comparison argument, we conclude that −M ≤ −v(x, t) ≤ uε (x, t) ≤ v(x, t) ≤ M, for every (x, t) ∈ Ω × [0, T ]. This ends the proof.

¤

teo.conv.J

Finally, we prove our last result, Theorem 54. teo.conv.J

lema.3.2.II

Proof of Theorem 54. By Lemma 55 we have that Fε (x, t) * 0 as measures in Ω × [0, T ] as ε → 0. Assume first that ψ ∈ C02+α (Ω) and let ϕ˜ε be the solution to wt − Lε w = 0 with w(x, 0) = ψ(x). Let ϕ˜ be a solution to

 ϕt − ∆ϕ = 0     ∂ϕ =0  ∂η    ϕ(x, 0) = ψ(x).

zero

Then, by Theorem 48 we know that ϕ˜ε → ϕ˜ uniformly in Ω × [0, T ]. For a fixed t > 0 set ϕε (x, s) = ϕ˜ε (x, t − s). Then ϕε satisfies ϕs + Lε ϕ = 0, ϕ(x, t) = ψ(x).

for s < t,


71

Analogously, set ϕ(x, s) = ϕ(x, ˜ t − s). Then ϕ satisfies  ϕt + ∆ϕ = 0     ∂ϕ =0  ∂η    ϕ(x, t) = ψ(x). Then, for wε = uε − u we have Z Z tZ Z tZ ∂ϕε ∂wε ε (x, s) ϕε (x, s) dx ds + (x, s) wε (x, s) dx ds w (x, t) ψ(x) dx = Ω 0 Ω ∂s 0 Ω ∂s Z tZ Z tZ = Lε (wε )(x, s)ϕε (x, s) dx ds + Fε (x, s) ϕε (x, s) dx ds 0 Ω 0 Ω Z tZ ∂ϕε + (x, s) wε (x, s) dx ds 0 Ω ∂s Z tZ Z tZ ε Fε (x, s) ϕε (x, s) dx ds Lε (ϕε )(x, s)w (x, s) dx ds + = 0 Ω 0 Ω Z tZ ∂ϕε + (x, s) wε (x, s) dx ds 0 Ω ∂s Z tZ = Fε (x, s)ϕε (x, s) dx ds. 0

Ω

lema.3.2.II

Now we observe that, by the Lemma 55, ¯Z t Z ¯ ¯Z t Z ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ F (x, s)ϕ (x, s) dx ds ≤ F (x, s)ϕ(x, s) dx ds ε ε ε ¯ ¯ ¯ ¯ 0 Ω 0 Ω Z tZ + sup kϕε (x, s) − ϕ(x, s)kL∞ (Ω) |Fε (x, s)| dx ds → 0 0 0,     u(x, 0) = u0 (x), x ∈ Ω.

In this model we prescribe the values of u outside Ω which is the analogous of prescribing the so called Dirichlet boundary conditions for the classical heat equation. However, the como.se.toma.el.dato.D boundary data is not understood in the usual sense as we will see in Remark 17 below. RAs explained before in this model the right hand side models the diffusion, the integral J(x−y)(u(y, t)−u(x, t)) dy takes into account the individuals arriving or leaving position x ∈ Ω from or to other places while we are prescribing the values of u outside the domain Ω by imposing u = g for x 6∈ Ω. When g = 0 we get that any individuals that leave Ω die, this is the case when Ω is surrounded by a hostile environment. 11.Dir

Existence and uniqueness of solutions of (6.1) is proved by a fixed point argument. Also a comparison principle is obtained. 0.16. Existence, uniqueness and a comparison principle. Existence and uniqueness of solutions is a consequence of Banach’s fixed point theorem. 11.Dir We look for u ∈ C([0, ∞); L1 (Ω)) satisfying (6.1). Fix t0 > 0 and consider the Banach space Xt0 = {w ∈ C([0, t0 ]; L1 (Ω))} with the norm |||w||| = max kw(·, t)kL1 (Ω) . 0≤t≤t0

We will obtain the solution as a fixed point of the operator T : Xt0 → Xt0 defined by Z tZ Tw0 (w)(x, t) = w0 (x) + J (x − y) (w(y, s) − w(x, s)) dy ds, 0

RN

where we impose w(x, t) = g(x, t), 73

for x 6∈ Ω.

74

on.Dir.lema

6. A LINEAR DIRICHLET PROBLEM WITH A RESCALE OF THE KERNEL

Lemma 57. Let w0 , z0 ∈ L1 (Ω), then there exists a constant C depending on J and Ω such that |||Tw0 (w) − Tz0 (z)||| ≤ Ct0 |||w − z||| + ||w0 − z0 ||L1 (Ω) for all w, z ∈ Xt0 . Proof. We have Z Z |Tw0 (w)(x, t) − Tz0 (z)(x, t)| dx ≤ |w0 − z0 |(x) dx Ω Ω Z ¯Z t Z h ¯ + ¯¯ J (x − y) (w(y, s) − z(y, s)) Ω 0 RN ¯ i ¯ −(w(x, s) − z(x, s)) dy ds¯ dx. Hence, taking into account that w and z vanish outside Ω, |||Tw0 (w) − Tz0 (z)||| ≤ ||w0 − z0 ||L1 (Ω) + Ct0 |||w − z|||, as we wanted to prove.

o.L^1.Dir.2

¤

Theorem 58. For every u0 ∈ L1 (Ω) there exists a unique solution u, such that u ∈ C([0, ∞); L1 (Ω)). Proof. We check first that Tu0 maps Xt0 into Xt0 . Taking z0 ≡ 0 and z ≡ 0 in Lemma contraction.Dir.lema 57 we get that Tu0 (w) ∈ C([0, t0 ]; L1 (Ω)) for any w ∈ Xt0 . contraction.Dir.lema Choose t0 such that Ct0 < 1. Now taking z0 ≡ w0 ≡ u0 in Lemma 57 we get that Tu0 is a strict contraction in Xt0 and the existence and uniqueness part of the theorem follows from Banach’s fixed point theorem in the interval [0, t0 ]. To extend the solution to [0, ∞) we may take as initial data u(x, t0 ) ∈ L1 (Ω) and obtain a solution up to [0, 2t0 ]. Iterating this procedure we get a solution defined in [0, ∞). ¤

Dirichlet.2

Remark 59. Note that in general a solution u with u0 > 0 and g = 0 is strictly positive in Ω (with a positive continuous extension to Ω) and vanishes outside Ω. Therefore a discontinuity occurs on ∂Ω and the boundary value is not taken in the usual ”classical” ChChR sense, see [34]. We now define what we understand by sub and supersolutions. 11.Dir

Super

Definition 60. A function u ∈ C([0, T ); L1 ((Ω)) is a supersolution of (6.1) if Z    u (x, t) ≥ J(x − y)(u(y, t) − u(x, t))dy, x ∈ Ω, t > 0,   t RN (6.2) u(x, t) ≥ g(x, t), x 6∈ Ω, t > 0,     u(x, 0) ≥ u0 (x), x ∈ Ω. As usual, subsolutions are defined analogously by reversing the inequalities.


mpar.0.lema

75

11.Dir

Lemma 61. Let u0 ∈ C(Ω), u0 ≥ 0, and u ∈ C(Ω × [0, T ]) a supersolution to (6.1) with g ≥ 0. Then, u ≥ 0. Proof. Assume for contradiction that u(x, t) is negative somewhere. Let v(x, t) = u(x, t) + εt with ε so small such that v is still negative somewhere. Then, if (x0 , t0 ) is a point where v attains its negative minimum, there holds that t0 > 0 and Z vt (x0 , t0 ) = ut (x0 , t0 ) + ε > J(x − y)(u(y, t0 ) − u(x0 , t0 )) dy RN Z = J(x − y)(v(y, t0 ) − v(x0 , t0 )) dy ≥ 0 RN

which is a contradiction. Thus, u ≥ 0.

¤

Corollary 62. Let J ∈ L∞ (RN ). Let u0 and v0 in L1 (Ω) with u ≥ v0 and g, h ∈ 11.Dir 0 L ((0, T ); L1 (RN \ Ω)) with g ≥ h. Let 11.Dir u be a solution of (6.1) with u(x, 0) = u0 and Dirichlet datum g and v be a solution of (6.1) with v(x, 0) = v0 and datum h. Then, u ≥ v a.e. ∞

Proof. Let w = u − v. Then, w is a supersolution with initial datum u0 − v0 ≥ 0 and datum g − h ≥ 0. Using the continuity of solutions with respect to the data and the fact compar.0.lema that J ∈ L∞ (RN ), we may assume that u, v ∈ C(Ω × [0, T ]). By Lemma 61 we obtain that w = u − v ≥ 0. So the corollary is proved. ¤ Corollary 63. Let u ∈ C(Ω × [0, T ]) (resp. v) be a supersolution (resp. subsolution) 11.Dir of (6.1). Then, u ≥ v. Proof. It follows the lines of the proof of the previous corollary.

¤

0.17. Convergence to the heat equation when rescaling the kernel.

calor.I.2

Let us consider the classical Dirichlet problem for the heat equation,   x ∈ Ω, t > 0, vt (x, t) − ∆v(x, t) = 0, (6.3) v(x, t) = g(x, t), x ∈ ∂Ω, t > 0,  v(x, 0) = u (x), x ∈ Ω. 0 11.Dir

calor.I.2

The nonlocal Dirichlet model (6.1) and the classical Dirichlet problem (6.3) share many properties, among them the asymptotic behavior of their solutions as t → ∞ is similar as ChChR was proved in [34].

calor.I.2

The main goal now is to show that the Dirichlet problem for the heat equation (6.3) 11.Dir can be approximated by suitable nonlocal problems of the form of (6.1).

rescales.2

More precisely, for a given J and a given ε > 0 we consider the rescaled kernel µ ¶ Z 1 ξ 1 2 −1 dz. (6.4) Jε (ξ) = C1 N J J(z)zN , with C1 = ε ε 2 B(0,d)

76

umann.C.1.2


Here C1 is a normalizing constant in order to obtain the Laplacian in the limit instead of a multiple of it. Let uε (x, t) be the solution of Z  Jε (x − y) ε ε   ut (x, t) = (u (y, t) − uε (x, t))dy, x ∈ Ω, t > 0,  2  ε Ω (6.5) u(x, t) = g(x, t), x 6∈ Ω, t > 0,     u(x, 0) = u0 (x), x ∈ Ω. Our main result now reads as follows.

teo.zero

ia.uniforme

C.1.u.tilde

umann.C.1.w

Theorem 64. Let Ω be a bounded C 2+α domain for some 0 < α < 1. calor.I.2

Neumann.C.1.2

Let v ∈ C 2+α,1+α/2 (Ω × [0, T ]) be the solution to (6.3) and let uε be the solution to (6.5) with Jε as above. Then, there exists C = C(T ) such that (6.6)

sup kv − uε kL∞ (Ω) ≤ Cεα → 0,

as ε → 0.

t∈[0,T ]

Related results for the Neumann problem where presented in the previous chapter (see CERW2 also [43]). Note that the assumed regularity of v is a consequence of regularity assumptions on Friedman the boundary data g, the domain Ω and the initial condition u0 , see [62]. teo.zero

In order to prove Theorem 64 let v˜ be a C 2+α,1+α/2 extension of v to RN × [0, T ]. Let us define the operator ˜ ε (z) = 1 L ε2 Then v˜ verifies (6.7)

Z

¡ ¢ Jε (x − y) z(y, t) − z(x, t) dy. RN

 ˜ ε (˜  v˜t (x, t) = L v )(x, t) + Fε (x, t)    v˜(x, t) = g(x, t) + G(x, t),     v˜(x, 0) = u (x), 0

x ∈ Ω, (0, T ], x 6∈ Ω, (0, T ], x ∈ Ω.

where, since ∆v = ∆˜ v in Ω, ˜ ε (˜ Fε (x, t) = −L v )(x, t) + ∆˜ v (x, t). Moreover as G is smooth and G(x, t) = 0 if x ∈ ∂Ω we have G(x, t) = O(ε),

for x such that dist(x, ∂Ω) ≤ εd.

We set wε = v˜ − uε and we note that  ˜ ε (wε )(x, t) + Fε (x, t)  wtε (x, t) = L    (6.8) wε (x, t) = G(x, t),     wε (x, 0) = 0,

x ∈ Ω, (0, T ], x 6∈ Ω, (0, T ], x ∈ Ω.


est1.2

77

First, we claim that, by the choice of C1 , the fact that J is radially symmetric and u˜ ∈ C 2+α,1+α/2 (RN × [0, T ]), we have that ˜ ε (˜ (6.9) sup kFε kL∞ (Ω) = sup k∆˜ v−L v )kL∞ (Ω) = O(εα ) . t∈[0,T ]

t∈[0,T ]

In fact,

¶ x−y ∆˜ v (x, t) − N +2 J (˜ v (y, t) − v˜(x, t)) dy ε ε RN becomes, under the change variables z = (x − y)/ε, Z C1 ∆˜ v (x, t) − 2 J (z) (˜ v (x − εz, t) − v˜(x, t)) dz ε RN est1.2 and hence (6.9) follows by a simple Taylor expansion. This proves the claim. C1

Z

µ

teo.zero

We proceed now to prove Theorem 64. teo.zero

Proof of Theorem 64. In order to prove the theorem by a comparisonwe first look for a supersolution. Let w be given by super en.omega

w(x, t) = K1 εα t + K2 ε.

(6.10)

For x ∈ Ω we have, if K1 is large, ˜ ˜ ε (wε )(x, t). (6.11) wt (x, t) − L(w)(x, t) = K1 εα ≥ Fε (x, t) = wtε (x, t) − L Since Gε (x, t) = O(ε) choosing K2 large, we obtain

fuera.omega

for x such that dist(x, ∂Ω) ≤ ε w(x, t) ≥ wε (x, t)

(6.12)

for x 6∈ Ω such that dist(x, ∂Ω) ≤ εd and t ∈ [0, T ]. Moreover it is clear that

ato.inicial

(6.13)

w(x, 0) = K2 ε > wε (x, 0) = 0.

en.omegafuera.omegadato.inicial

Thanks to (6.11), (6.12) and (6.13) we can apply the comparison result and conclude that

cota.arriba

(6.14)

wε (x, t) ≤ w(x, t) = K1 εα t + K2 ε.

In a similar fashion we prove that w(x, t) = −K1 εα t − K2 ε is a subsolution and hence cota.abajo

(6.15)

wε (x, t) ≥ w(x, t) = −K1 εα t − K2 ε.

Therefore final

(6.16)

sup ku − uε kL∞ (Ω) ≤ C(T )εα , t∈[0,T ]

as we wanted to prove.

¤

CHAPTER 7

Scaling the kernel in a higher order problem

ling.higher αn

2 v(x, t) can be In this short chapter we show that the problem vt (x, t) = −An (−∆) sec.linear.higher approximated by nonlocal problems like the one presented in Chapter 4 when rescaled in an appropriate way.

nonloceq.2

teo.escalas

nonloceq.ve

Let us recall that we have considered  ut (x, t) = (−1)n−1 (J ∗ Id − 1)n (u(x, t))   Ã !  n µ ¶  X n = (−1)n−1 (−1)n−k (J∗)k (u) (x, t), (7.1)  k  k=0   u(x, 0) = u0 (x), for x ∈ RN and t > 0, as a nonlocal higher order equation. Theorem 65. Let uε be the unique solution to ( (7.2)

(uε )t (x, t) = (−1)n−1 u(x, 0) = u0 (x),

(Jε ∗ Id − 1))n (uε (x, t)), εαn

where Jε (s) = ε−N J( sε ). Then, for every T > 0, we have lim kuε − vkL∞ (RN ×(0,T )) = 0,

ε→0

where v is the solution to the local problem vt (x, t) = −An (−∆) initial condition v(x, 0) = u0 (x). teo.escalas

αn 2

v(x, t) with the same

Proof of Theorem 65. The proof uses once more the explicit formula for the solutions in Fourier variables. We have, arguing exactly as before, n c ε (ξ)−1) n−1 (J t εαn

uˆε (ξ, t) = e(−1)

uˆ0 (ξ).

and vˆ(ξ, t) = e−A

n |ξ|αn t

79

uˆ0 (ξ).

80

7. SCALING THE KERNEL IN A HIGHER ORDER PROBLEM

ˆ Now, we just observe that Jbε (ξ) = J(εξ) and therefore we obtain Z Z ¯ ¯ n ˆ ¯ (−1)n−1 (J(εξ)−1) ¯ t −An |ξ|αn t αn ε |uˆε − vˆ| (ξ, t) dξ = −e )uˆ0 (ξ)¯ dξ ¯(e RN µZ ¯ RN ¯ n ˆ ¯ (−1)n−1 (J(εξ)−1) t −An |ξ|αn t ¯ αn ε ≤ kuˆ0 kL∞ (RN ) −e ¯e ¯ dξ |ξ|≥r(ε) Z ¯ ¯ ¶ n ˆ n |ξ|αn t ¯ ¯ (−1)n−1 (J(εξ)−1) t −A αn ε + −e ¯e ¯ dξ . |ξ| 0, x ∈ RN , (8.1) u(0, x) = u0 (x), x ∈ RN . In this paper we analyze a nonlocal equation that takes into account convective and diffusive effects. We deal with the nonlocal evolution equation ( ut (x, t) = (J ∗ u − u) (x, t) + (G ∗ (f (u)) − f (u)) (x, t), t > 0, x ∈ Rd , (8.2) u(0, x) = u0 (x), x ∈ Rd . LetRus state first Rour basic assumptions. The functions J and G are nonnegatives and verify Rd J(x)dx = Rd G(x)dx = 1. Moreover, we consider J smooth, J ∈ S(Rd ), the space of rapidly decreasing functions, and radially symmetric and G smooth, G ∈ S(Rd ), but not necessarily symmetric. To obtain a diffusion operator similar to the Laplacian we impose in addition that J verifies Z 1 2 b 1 J(z)zi2 dz = 1. ∂ J(0) = 2 ξi ξi 2 supp(J) This implies that b − 1 + ξ 2 ∼ |ξ|3 , J(ξ)

for ξ close to 0.

Here Jb is the Fourier transform of J and the notation A ∼ B means that there exist constants C1 and C2 such that C1 A ≤ B ≤ C2 A. We can consider more general kernels J b − 1 + A ξ 2 ∼ |ξ|3 . Since the results with expansions in Fourier variables of the form J(ξ) (and the proofs) are almost the same, we do not include the details for this more general case, but we comment on how the results are modified by the appearance of A. The nonlinearity f will de assumed nondecreasing with f (0) = 0 and locally Lipschitz continuous (a typical example that we will consider below is f (u) = |u|q−1 u with q > 1). nonlocal.2 In our case, see the equation in (8.2), we have a diffusion operator J ∗ u − u and a nonlinear convective part given by G ∗ (f (u)) − f (u). Concerning this last term, if G is not symmetric then individuals have greater probability of jumping in one direction than in others, provoking a convective effect. 81

o.existence

usion.local

82

8. A NON-LOCAL CONVECTION DIFFUSION EQUATION

nonlocal.2

We will call equation (8.2), a nonlocal convection-diffusion equation. It is nonlocal since the diffusion of the density u at a point x and time t does not only depend on u(x, t) and its derivatives at that point (x, t), but on all the values of u in a fixed spatial neighborhood of x through the convolution terms J ∗ u and G ∗ (f (u)) (this neighborhood depends on the supports of J and G). First, we prove existence, uniqueness and well-possedness of a solution with an initial condition u(0, x) = u0 (x) ∈ L1 (Rd ) ∩ L∞ (Rd ). Theorem 66. For any u0 ∈ L1 (Rd ) ∩ L∞ (Rd ) there exists a unique global solution u ∈ C([0, ∞); L1 (Rd )) ∩ L∞ ([0, ∞); Rd ). nonlocal.2

If u and v are solutions of (8.2) corresponding to initial data u0 , v0 ∈ L1 (Rd ) ∩ L∞ (Rd ) respectively, then the following contraction property ku(t) − v(t)kL1 (Rd ) ≤ ku0 − v0 kL1 (Rd ) holds for any t ≥ 0. In addition, ku(t)kL∞ (Rd ) ≤ ku0 kL∞ (Rd ) . We haveChChR to emphasize that a lack of regularizing effectEZoccurs. This has been already observed in [34] for the linear problem wt = J ∗ w − w. In [54], the authors prove that the solutions to the local convection-diffusion problem, ut = ∆u + b · ∇f (u), satisfy an estimate of the form ku(t)kL∞ (Rd ) ≤ C(ku0 kL1 (Rd ) ) t−d/2 for any initial data u0 ∈ L1 (Rd )∩L∞ (Rd ). In our nonlocal model, we cannot prove such type of inequality independently of the L∞ (Rd )norm of the initial data. Moreover, in the one-dimensional case with a suitable bound on the nonlinearity that appears in the convective part, f , we can prove that such an inequality does not hold in general, see Chapter 3. In addition, the L1 (Rd ) − L∞ (Rd ) regularizing effect is not available for the linear equation, wt = J ∗ w − w. When J is nonnegative and compactly supported, the equation wt = J ∗ w − w shares many properties with the classical heat equation, wt = ∆w, such as: bounded stationary solutions are constant, a maximum principle holds for both of them and perturbations F propagate with infinite speed, see [59]. However, there is no regularizing effect in general. CERW CERW2 Moreover, in [42] and [43] nonlocal Neumann boundary conditions where taken into account. It is proved there that solutions of the nonlocal problems converge to solutions of the heat equation with Neumann boundary conditions when a rescaling parameter goes to zero. nonlocal.2

Concerning (8.2) we can obtain a solution to a standard convection-diffusion equation (8.3)

vt (x, t) = ∆v(x, t) + b · ∇f (v)(x, t), nonlocal.2

t > 0, x ∈ Rd ,

as the limit of solutions to (8.2) when a scaling parameter goes to zero. In fact, let us consider 1 ³s´ 1 ³s´ Jε (s) = d J , Gε (s) = d G , ε ε ε ε


Rescaladas

.ve.to.zero

83

and the solution uε (x, t) to our convection-diffusion problem rescaled adequately, Z  1   (u ) (x, t) = 2 Jε (x − y)(uε (y, t) − uε (x, t)) dy   ε t ε Rd Z 1 (8.4) + Gε (x − y)(f (uε (y, t)) − f (uε (x, t))) dy,   ε Rd   uε (x, 0) = u0 (x). Remark that the scaling is different for the diffusive part of the equation J ∗ u − u and for the convective part G ∗ f (u) − f (u). The same different scaling properties can be observed for the local terms ∆u and b · ∇f (u). Theorem 67. With the above notations, for any T > 0, we have lim sup kuε − vkL2 (Rd ) = 0,

ε→0 t∈[0,T ]

convection-diffusion.

where v(x, t) is the unique solution to the local convection-diffusion problem (8.3) with initial condition v(x, 0) = u0 (x) ∈ L1 (Rd ) ∩ L∞ (Rd ) and b = (b1 , ..., bd ) given by Z bj = xj G(x) dx, j = 1, ..., d. Rd

This result justifies the use of the name “nonlocal convection-diffusion problem” when nonlocal.2 we refer to (8.2). b − 1 − ib · ξ| ≤ C|ξ|2 From our hypotheses on J and G it follows that they verify |G(ξ) b − 1 + ξ 2 | ≤ C|ξ|3 for every ξ ∈ Rd . These bounds are exactly what we are using and |J(ξ) in the proof of this convergence result. Remark that when G is symmetric then b = 0 and we obtain the heat equation in the limit. Of course the most interesting case is when b 6= 0 (this happens when G is not symmetric). Also we note that the conclusion of the theorem holds for other Lp (Rd )-norms besides L2 (Rd ), however the proof is more involved. We can consider kernels J such that Z 1 A= J(z)zi2 dz 6= 1. 2 supp(J) b This gives the expansion J(ξ)−1+ Aξ 2 ∼ |ξ|3 , for ξ close to 0. In this case we will arrive to a convection-diffusion equation with a multiple of the Laplacian as the diffusion operator, vt = A∆v + b · ∇f (v).

nonlocal.2

Next, we want to study the asymptotic behaviour as t → ∞ of solutions to (8.2). To this end we first analyze the decay of solutions taking into account only the diffusive part (the linear part) of the equation.ChChR These solutions have a similar decay rate as the one that liv.rossi holds for the heat equation, see [34] and [68] where the Fourier transform play a key role. Using similar techniques we can prove the following result that deals with this asymptotic decay rate.

84

th.decay

inear.intro


Theorem 68. Let p ∈ [1, ∞]. For any u0 ∈ L1 (Rd ) ∩ L∞ (Rd ) the solution w(x, t) of the linear problem ( wt (x, t) = (J ∗ w − w)(x, t), t > 0, x ∈ Rd , (8.5) u(0, x) = u0 (x), x ∈ Rd , satisfies the decay estimate d

1

kw(t)kLp (Rd ) ≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) hti− 2 (1− p ) . Throughout this paper we will use the notation A ≤ hti−α to denote A ≤ (1 + t)−α . Now we are ready to face the study of the asymptotic behaviour of the complete probnonlocal.2 lem (8.2). To this end we have to impose some grow condition on f . Therefore, in the sequel we restrict ourselves to nonlinearities f that are pure powers grow.f

(8.6)

f (u) = |u|q−1 u

with q > 1. The analysis is more involved than the one performed for the linear part and we cannot use here the Fourier transform directly (of course, by the presence of the nonlinear term). Our strategy is to write a variation of constants formula for the solution and then prove estimates that say that the nonlinear part decay faster than the linear one. For the local EZ convection diffusion equation this analysis was performed by Escobedo and Zuazua in [54]. However, in the previously mentioned reference energy estimates were used together with Sobolev inequalities to obtain decay bounds. These Sobolev inequalities are not available for the nonlocal model, since the linear part does not have any regularizing effect, see remark.energy Remark 87. Therefore, we have to avoid the use of energy estimates and tackle the problem using a variantMR571048 of the Fourier splitting method proposed by Schonbek to deal with local MR856876 MR1356749 problems, see [78], [79] and [80]. We state our result concerning the asymptotic behaviour (decay rate) of the complete nonlocal model as follows: long.time

decay.rate

grow.f

Theorem 69. Let f satisfies (8.6) with nonlocal.2 q > 1 and u0 ∈ L1 (Rd ) ∩ L∞ (Rd ). Then, for every p ∈ [1, ∞) the solution u of equation (8.2) verifies (8.7)

d

1

ku(t)kLp (Rd ) ≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) hti− 2 (1− p ) .

Finally, we look at the first order term in the asymptotic expansion of the solution. For q > (d + 1)/d, we find that this leading order term is the same as the one that appears in the linear local heat equation. This is due to the fact that the nonlinear convection is of higher order and thatChChR the radially symmetric diffusion leads to gaussian kernels in the liv.rossi asymptotic regime, see [34] and [68].


ak.behavior

85

grow.f

Theorem 70. Let f satisfies (8.6) with q > (d + 1)/d and let the initial condition u0 belongs to L1 (Rd , 1 + |x|) ∩ L∞ (Rd ). For any p ∈ [2, ∞) the following holds 1

d

t− 2 (1− p ) ku(t) − M H(t)kLp (Rd ) ≤ C(J, G, p, d) αq (t), where

Z u0 (x) dx,

M= Rd

H(t) is the Gaussian, x2

H(t) = and

( αq (t) =

1

hti− 2 hti

1−d(q−1) 2

e− 4t

d

(2πt) 2

,

if q ≥ (d + 2)/d, if (d + 1)/d < q < (d + 2)/d.

Remark that we prove a weak nonlinear behaviour, in fact the decay rate and the first order term in the expansion are the same that appear in the linear model wt = J ∗ w − w, liv.rossi see [68]. b − (1 − |ξ|2 ) ∼ B|ξ|3 , for ξ As before, recall that our hypotheses on J imply that J(ξ) teo.weak.behavior close to 0. This is the key property of J used in the proof of Theorem 70. We note that b − (1 − A|ξ|2 ) ∼ B|ξ|3 , for ξ ∼ 0, we get as when we have an expansion of the form J(ξ) first order term a Gaussian profile of the form HA (t) = H(At). Also note that q = (d + 1)/dEZis a critical exponent for the local convection-diffusion problem, vt = ∆v + b · ∇(v q ), see [54]. When q is supercritical, q > (d + 1)/d, for the local equation it also holds an asymptotic simplification to the heat semigroup as t → ∞. The first order term in the asymptotic behaviour for critical or subcritical exponents 1 < q ≤ (d + 1)/d is left open. One of the main difficulties that one has to face here is the absence of a self-similar profile due to the inhomogeneous behaviour of the convolution kernels.

linear

0.18. The linear semigroup. In this chapter we analyze the asymptotic behavior of the solutions of the equation ( wt (x, t) = (J ∗ w − w)(x, t), t > 0, x ∈ Rd , (8.8) w(0, x) = u0 (x), x ∈ Rd . As we have mentioned in the introduction, when J is nonnegative and compactly supported, this equation shares many properties with the classical heat equation, wt = ∆w, such as: bounded stationary solutions are constant, a maximum principle holds for both F of them and perturbations propagate with infinite speed, see [59]. However, there is no

86


regularizing effect in general. In fact, the singularity of the source solution, that is a solulinear tion to (8.8) with initial condition a delta measure, u0 = δ0 , remains with an exponential decay. In fact, this fundamental solution can be decomposed as S(x, t) = e−t δ0 + Kt (x) sol.fundamental where Kt (x) is smooth, see Lemma 5. In this way we see that there is no regularizing effect linear since the solution w of (8.8) can be written as w(t) = S(t) ∗ u0 = e−t u0 + Kt ∗ u0 with Klinear t smooth, which means that w(·, t) is as regular as u0 is. This fact makes the analysis of (8.8) more involved.

mental.lema decomp.2

linear

linear

Lemma 71. The fundamental solution of (8.8), that is the solution of (8.8) with initial condition u0 = δ0 , can be decomposed as S(x, t) = e−t δ0 (x) + Kt (x),

(8.9)

linear

with Kt (x) = K(x, t) smooth. Moreover, if u is the solution of (8.8) it can be written as Z w(x, t) = (S ∗ u0 )(x, t) = S(x, t − y)u0 (y) dy. R

linear

Proof. Applying the Fourier transform to (8.8) we obtain that b − 1). w bt (ξ, t) = w(ξ, b t)(J(ξ) Hence, as the initial datum verifies ub0 = δb0 = 1, b

b

w(ξ, b t) = e(J(ξ)−1)t = e−t + e−t (eJ(ξ)t − 1). The first part of the lemma follows applying the inverse Fourier transform. linear

To finish the proof we just observe that S ∗ u0 is a solution of (8.8) (just use Fubini’s theorem) with (S ∗ u0 )(0, x) = u0 (x). ¤

fund.sol

In the following we will give estimates on the regular part of the fundamental solution Kt defined by: Z b (8.10) Kt (x) = (et(J(ξ)−1) − e−t ) eix·ξ dξ, Rd

that is, in the Fourier space, b b t (ξ) = et(J(ξ)−1) K − e−t .

The behavior of Lp (Rd )-norms of Kt will be obtained by analyzing the cases p = ∞ and p = 1. The case p = ∞ follows by Hausdorff-Young’s inequality. The case p = 1 follows linear by using the fact that the L1 (Rd )-norm of the solutions to (8.8) does not increase. The analysis of the behaviour of the gradient ∇Kt is more involved. The behavior of Lp (Rd )-norms with 2 ≤ p ≤ ∞ follows by Hausdorff-Young’s inequality in the case p = ∞ to and Plancherel’s identity for p = 2. However, the case 1 ≤ p < 2 is more tricky. In order MR0461121 evaluate the L1 (Rd )-norm of ∇Kt we will use the Carlson inequality (see for instance [25]) carlson

(8.11)

1−

d

d

m 2m 2m kϕkL1 (Rd ) ≤ C kϕkL2 (R d ) k|x| ϕkL2 (R) ,


87

which holds for m > d/2. The use of the above inequality with ϕ = ∇Kt imposes that |x|m ∇Kt belongs to L2 (Rd ). To guarantee this property and to obtain the decay rate for nucleu the L2 (Rd )-norm of |x|m ∇Kt we will use in Lemma 74 that J ∈ S(Rd ). The following lemma gives us the decay rate of the Lp (Rd )-norms of the kernel Kt for 1≤p≤∞. decay.kt

b ∈ L1 (Rd ), ∂ξ J(ξ) b ∈ L2 (Rd ) and Lemma 72. Let J be such that J(ξ) b − 1 + ξ 2 ∼ |ξ|3 , J(ξ)

b ∼ −ξ ∂ξ J(ξ)

as ξ ∼ 0.

fund.sol

For any p ≥ 1 there exists a positive constant c(p, J) such that Kt , defined in (8.10), satisfies: lp.decay

d

1

kKt kLp (Rd ) ≤ c(p, J) hti− 2 (1− p )

(8.12) for any t > 0.

Remark 73. In fact, when p = ∞, a stronger inequality can be proven, b L1 (Rd ) + C hti−d/2 , kKt kL∞ (Rd ) ≤ Cte−δt kJk for some positive δ = δ(J). Moreover, for p = 1 we have, kKt kL1 (Rd ) ≤ 2 and for any p ∈ [1, ∞] kS(t)kLp (Rd )−Lp (Rd ) ≤ 3. decay.kt

Proof of Lemma 72. We analyze the cases p = ∞ and p = 1, the others can be easily obtained applying Hölder’s inequality. Case p = ∞. Using Hausdorff-Young’s inequality we obtain that Z b kKt kL∞ (Rd ) ≤ |et(J(ξ)−1) − e−t |dξ. Rd

cond1

cond2

Observe that the symmetry of J guarantees that Jb is a real number. Let us choose R > 0 such that |ξ|2 b (8.13) |J(ξ)| ≤1− for all |ξ| ≤ R. 2 Once R is fixed, there exists δ = δ(J), 0 < δ < 1, with b (8.14) |J(ξ)| ≤ 1 − δ for all |ξ| ≥ R. Using that for any real numbers a and b the following inequality holds: |ea − eb | ≤ |a − b| max{ea , eb } we obtain for any |ξ| ≥ R,

est.on.kt

(8.15)

b b b b |et(J(ξ)−1) − e−t | ≤ t|J(ξ)| max{e−t , et(J(ξ)−1) } ≤ te−δt |J(ξ)|.

88


Then the following integral decays exponentially, Z Z b t(J(ξ)−1) −t −δt |e − e |dξ ≤ e t |ξ|≥R

est.6

b |J(ξ)|dξ. |ξ|≥R

Using that this term is exponentially small, it remains to prove that Z b (8.16) I(t) = |et(J(ξ)−1) − e−t |dξ ≤ Chti−d/2 . |ξ|≤R

To handle this case we use the following estimates: Z Z b t(J(ξ)−1) −t |I(t)| ≤ e dξ + e C(R) ≤ |ξ|≤R

and

dξ + e−t C(R) ≤ C(R) |ξ|≤R

Z

Z b t(J(ξ)−1)

e |ξ|≤R Z

|I(t)| ≤

−t

e−

dξ + e C(R) ≤

t|ξ|2 2

+ e−t C(R)

|ξ|≤R |η|2 − 2

= t−d/2

e

+ e−t C(R) ≤ Ct−d/2 .

|η|≤Rt1/2

est.6

The last two estimates prove (8.16) and this finishes the analysis of this case.

linear.intro

1 Case p = 1. First we prove that the Llinear.intro (Rd )-norm of the solutions to equation (8.5) does not increase. Multiplying equation (8.5) by sgn(w(x, t)) and integrating in space variable we obtain, Z Z Z Z d |w(x, t)| dx = J(x − y)w(y, t)sgn(w(x, t)) dy ds − |w(t, x)| dx dt Rd Rd Rd Rd Z Z Z ≤ J(x − y)|w(y, t)| dx dy − |w(x, t)| dx ≤ 0, Rd

Rd

Rd

which shows that the L1 (Rd )-norm does not increase. Hence, for any u0 ∈ L1 (Rd ), the following holds: Z Z −t |e u0 (x) + (Kt ∗ u0 )(x)| dx ≤ |u0 (x)| dx, Rd

and as a consequence,

Rd

Z

Z |(Kt ∗ u0 )(x)| dx ≤ 2

Rd

|u0 (x)| dx. Rd

Choosing (u0 )n ∈ L1 (Rd ) such that (u0 )n → δ0 in S 0 (Rd ) we obtain in the limit that Z |Kt (x)|dx ≤ 2. Rd

This ends the proof of the L1 -case and finishes the proof.

¤


89

The following lemma will play a key role when analyzing the decay of the complete nonlocal.2 problem (8.2). In the sequel we will denote by L1 (Rd , a(x)) the following space: ½ Z ¾ 1 d L (R , a(x)) = ϕ : a(x)|ϕ(x)|dx < ∞ . Rd

nucleu

Lemma 74. Let p ≥ 1 and J ∈ S(Rd ). There exists a positive constant c(p, J) such that d

1

1

kKt ∗ ϕ − Kt kLp (Rd ) ≤ c(p)hti− 2 (1− p )− 2 kϕkL1 (Rd ,|x|) holds for all ϕ ∈ L1 (Rd , 1 + |x|).

est.10

Proof. Explicit computations shows that Z Z (Kt ∗ ϕ − Kt )(x) = Kt (x − y)ϕ(y)dy − Kt (x) dx d Rd ZR = ϕ(y)(Kt (x − y) − Kt (x)) dy Rd Z Z 1 (8.17) = ϕ(y) ∇Kt (x − sy) · (−y) ds dy. Rd

0

We will analyze the cases p = 1 and p = ∞, the others cases follow by interpolation. For p = ∞ we have, Z star1

(8.18)

kKt ∗ ϕ − Kt kL∞ (Rd ) ≤ k∇Kt kL∞ (Rd )

|y||ϕ(y)| dy. Rd

est.10

star2

In the case p = 1, by using (8.17) the following holds: Z Z Z Z 1 |(Kt ∗ ϕ − Kt )(x)| dx ≤ |y||ϕ(y)| |∇Kt (x − sy)| ds dy dx Rd Rd Rd 0 Z Z 1Z = |y||ϕ(y)| |∇Kt (x − sy)| dx ds dy Rd 0 Rd Z Z (8.19) = |y||ϕ(y)| dy |∇Kt (x)| dx. Rd

star1

Rd

star2

In view of (8.18) and (8.19) it is sufficient to prove that d

1

k∇Kt kL∞ (Rd ) ≤ Chti− 2 − 2 and 1

k∇Kt kL1 (Rd ) ≤ Chti− 2 .

90


cond1

cond2

Inest.on.kt the first case, with R and δ as in (8.13) and (8.14), by Hausdorff-Young’s inequality and (8.15) we obtain: Z b k∇Kt kL∞ (Rd ) ≤ |ξ||et(J(ξ)−1) − e−t |dξ d ZR Z b b t(J(ξ)−1) −t = |ξ||e − e |dξ + |ξ||et(J(ξ)−1) − e−t |dξ |ξ|≤R |ξ|≥R Z Z Z −t|ξ|2 /2 −t −δt b ≤ |ξ|e dξ + e |ξ|dξ + t |ξ||J(ξ)|e dξ |ξ|≤R

|ξ|≤R

≤ C(R)hti

− d2 − 12

+ C(R)e

− d2 − 12

,

≤ C(J) hti

−t

|ξ|≥R

+ C(J) t e

−δt

b belongs to L1 (Rd ). provided that |ξ|J(ξ) In the second case it is enough to prove that the L1 (Rd )-norm of ∂x1 Kt is controlled by hti−1/2 . In this case Carlson’s inequality gives us 1−

d

d

m 2m 2m k∂x1 Kt kL1 (Rd ) ≤ C k∂x1 Kt kL2 (R d ) k|x| ∂x1 Kt kL2 (Rd ) ,

for any m > d/2. Now our aim is to prove that, for any t > 0, we have kk1

d

1

k∂x1 Kt kL2 (Rd ) ≤ C(J)hti− 4 − 2

(8.20) and

kk2

k|x|m ∂x1 Kt kL2 (Rd ) ≤ C(J)hti

(8.21)

m−1 d −4 2

.

est.on.kt b belongs to L2 (Rd ) we obtain By Plancherel’s identity, estimate (8.15) and using that |ξ|J(ξ) Z b 2 k∂x1 Kt kL2 (Rd ) = |ξ1 |2 |et(J(ξ)−1) − e−t |2 dξ Rd Z Z Z 2 −t|ξ|2 −2t 2 b 2 dξ ≤ 2 |ξ1 | e dξ + e |ξ1 | dξ + |ξ1 |2 e−2δt t2 |J(ξ)| |ξ|≤R

≤ C(R)hti

|ξ|≤R − d2 − 12

+ C(R)e

− d2 − 12

.

≤ C(J) hti

−2t

|ξ|≥R −2δt 2

+ C(J)e

t

kk1

This shows (8.20). kk2 To prove (8.21), observe that Z k|x|

m

∂x1 Kt k2L2 (Rd )

≤ c(d) Rd

2 2m (x2m 1 + · · · + xd )|∂x1 Kt (x)| dx.


91

Thus, by symmetry it is sufficient to prove that Z b t (ξ))|2 dξ ≤ C(J) htim−1− d2 |∂ξm1 (ξ1 K Rd

and

Z d

Rd

b t (ξ))|2 dξ ≤ C(J) htim−1− 2 . |∂ξm2 (ξ1 K

Observe that b t (ξ))| = |ξ1 ∂ξm K b t (ξ) + m∂ m−1 K b t (ξ)| ≤ |ξ||∂ξm K b t (ξ)| + m|∂ m−1 K b t (ξ)| |∂ξm (ξ1 K 1

ξ1

1

ξ1

1

and

b t )| ≤ |ξ||∂ m K b |∂ξm2 (ξ1 K ξ2 t (ξ)|. Hence we just have to prove that Z b t (ξ)|2 dξ ≤ C(J) htin−r− d2 , |ξ|2r |∂ξn1 K (r, n) ∈ {(0, m − 1), (1, m)} . Rd

Choosing m = [d/2] + 1 (the notation [·] stands for the floor function) the above inequality has to hold for n = [d/2], [d/2] + 1. First we recall the following elementary identity X ∂ξn1 (eg ) = eg ai1 ,...,in (∂ξ11 g)i1 (∂ξ21 g)i2 ...(∂ξn1 g)in , i1 +2i2 +...+nin =n

where ai1 ,...,in are universal constants independent of g. Tacking into account that b ct (ξ) = et(J(ξ)−1) K − e−t

we obtain X

b

ct (ξ) = et(J(ξ)−1) ∂ξn1 K

ai1 ,...,in ti1 +···+in

i1 +2i2 +...+nin =n

n Y

b ij [∂ξj1 J(ξ)]

j=1

and hence ct (ξ)|2 |∂ξn1 K

≤ Ce

b 2t(J(ξ)−1)

X i1 +2i2 +...+nin =n

t

2(i1 +···+in )

n Y

b 2ij . [∂ξj1 J(ξ)]

j=1

Using that all the partial derivatives of Jb decay faster than any polinomial in |ξ|, as |ξ| → ∞, we obtain that Z ct (ξ)|2 dξ ≤ C(J) e−2δt hti2n |ξ|2r |∂ n K |ξ|>R

ξ1

cond1 cond2 b is smooth where R and δ are chosen as in (8.13) and (8.14). Tacking into account that J(ξ) (since J ∈ S(Rd )) we obtain that for all |ξ| ≤ R the following hold:

b |∂ξ1 J(ξ)| ≤ C |ξ|

92


and

b |∂ξj1 J(ξ)| ≤ C,

Then for all |ξ| ≤ R we have ct (ξ)|2 ≤ C e−t|ξ| |∂ξn1 K

2

j = 2, . . . , n. X

t2(i1 +···+in ) |ξ|2i1 .

i1 +2i2 +...+nin =n

Finally, using that for any l ≥ 0 Z d

2

l

e−t|ξ| |ξ|l dξ ≤ C(R)hti− 2 − 2 , |ξ|≤R

we obtain

Z

X

d

|ξ|≤R

|ξ|2r |∂ξn1 Kt (ξ)|2 dξ ≤ C(R)hti− 2

hti2p(i1 ,...,id )−r

i1 +2i2 +···+nin =n

where p(i1 , . . . , in ) = (i1 + · · · + in ) − =

i1 2

i1 i1 + 2i2 + . . . nin n + i2 + · · · + in ≤ = . 2 2 2


¤

We now prove a decay estimate that takes into account the linear semigroup applied to the convolution with a kernel G.

.term.deriv

Lemma 75. Let 1 ≤ p ≤ r ≤ ∞, J ∈ S(Rd ) and G ∈ L1 (Rd , |x|). There exists a positive constant C = C(p, J, G) such that the following estimate

derivat.new

(8.22)

d 1

1

1

kS(t) ∗ G ∗ ϕ − S(t) ∗ ϕkLr (Rd ) ≤ Chti− 2 ( p − r )− 2 (kϕkLp (Rd ) + kϕkLr (Rd ) ).

holds for all ϕ ∈ Lp (Rd ) ∩ Lr (Rd ). Remark 76. In fact the following stronger inequality holds: d 1

1

1

kS(t) ∗ G ∗ ϕ − S(t) ∗ ϕkLr (Rd ) ≤ C hti− 2 ( p − r )− 2 kϕkLp (Rd ) + C e−t kϕkLr (Rd ) . Proof. We write S(t) as S(t) = e−t δ0 + Kt and we get S(t) ∗ G ∗ ϕ − S(t) ∗ ϕ = e−t (G ∗ ϕ − ϕ) + Kt ∗ G ∗ ϕ − Kt ∗ ϕ. The first term in the above right hand side verifies: e−t kG ∗ ϕ − ϕkLr (Rd ) ≤ e−t (kGkL1 (Rd ) kϕkLr (Rd ) + kϕkLr (Rd ) ) ≤ 2e−t kϕkLr (Rd ) . nucleu

For the second one, by Lemma 74 we get that Kt satisfies d

1

1

kKt ∗ G − Kt kLa (Rd ) ≤ C(r, J)kGkL1 (Rd ,|x|) hti− 2 (1− a )− 2

for all t ≥ 0 where a is such that 1/r = 1/a + 1/p − 1. Then, using Young’s inequality we end the proof. ¤

tegral.form


93

0.19. Existence and uniqueness. In this chapter we use the previous results and estimates on the linear semigroup to prove the existence and uniqueness of the solution to nonlocal.2 our nonlinear problem (8.2). The proof is based on the variation of constants formula and uses the previous properties of the linear diffusion semigroup. teo.existence

Proof of Theorem 66. Recall that we want prove the global existence of solutions for initial conditions u0 ∈ L1 (Rd ) ∩ L∞ (Rd ). nonlocal.2

Let us consider the following integral equation associated with (8.2): Z t (8.23) u(t) = S(t) ∗ u0 + S(t − s) ∗ (G ∗ (f (u)) − f (u))(s) ds, 0

the functional Z

t

Φ[u](t) = S(t) ∗ u0 +

S(t − s) ∗ (G ∗ (f (u)) − f (u))(s) ds 0

and the space X(T ) = C([0, T ]; L1 (Rd )) ∩ L∞ ([0, T ]; Rd ) endowed with the norm ¡ ¢ kukX(T ) = sup ku(t)kL1 (Rd ) + ku(t)kL∞ (Rd ) . t∈[0,T ]

We will prove that Φ is a contraction in the ball of radius R, BR , of XT , if T is small enough. Step I. Local Existence. Let M = max{ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) } and p = 1, ∞. Then, decay.kt using the results of Lemma 72 we obtain, kΦ[u](t)kLp (Rd ) ≤ kS(t) ∗ u0 kLp (Rd ) Z t + kS(t − s) ∗ G ∗ (f (u)) − S(t − s) ∗ f (u)kLp (Rd ) ds 0 −t

≤ (e

Z

+ kKt kL1 (Rd ) )ku0 kLp (Rd ) t

+

2(e−(t−s) + kKt−s kL1 (Rd ) )kf (u)(s)kLp (Rd ) ds

0

≤ 3 ku0 kLp (Rd ) + 6 T f (R) ≤ 3M + 6 T f (R). This implies that kΦ[u]kX(T ) ≤ 6M + 12 T f (R). Choosing R = 12M and T such that 12 T f (R) < 6M we obtain that Φ(BR ) ⊂ BR .

94


Let us choose u and v in BR . Then for p = 1, ∞ the following hold: Z t kΦ[u](t) − Φ[v](t)kLp (Rd ) ≤ k(S(t − s) ∗ G − S(t − s)) ∗ (f (u) − f (v))kLp (Rd ) ds 0Z t ≤6 kf (u)(s) − f (v)(s)kLp (Rd ) ds 0 Z t ≤ C(R) ku(s) − v(s)kLp (Rd ) ds 0

≤ C(R) T ku − vkX(T ) . Choosing T small weintegral.form obtain that Φ[u] is a contraction in BR and then there exists a unique local solution u of (8.23). Step II. Global existence. To prove the global well posedness of the solutions we have to guarantee that both L1 (Rd ) and L∞ (Rd )-norms of the solutions do not blow up in finite time. We will apply the following lemma to control the L∞ (Rd )-norm of the solutions. lema.ineg

ineg.rara

eg.rara.neg

Lemma 77. Let θ ∈ L1 (Rd ) and K be a nonnegative function with mass one. Then for any µ ≥ 0 the following hold: Z Z Z (8.24) K(x − y)θ(y) dy dx ≤ θ(x) dx Rd

θ(x)>µ

and

Z

θ(x)>µ

Z

Z

(8.25)

K(x − y)θ(y) dy dx ≥ θ(x)0 θ(y)>0 Z Z = θ(y) K(x − y) dx dy θ(y)>0 θ(x)>0 Z Z ≤ θ(y) K(x − y) dx dy θ(y)>0 Rd Z = θ(y) dy. θ(y)>0


95

Now let us analyze the general case µ > 0. In this case the following inequality Z Z θ(x) dx ≤ |θ(x)| dx θ(x)>µ

Rd

shows that the set {x ∈ Rd : θ(x) > µ} has finite measure. Then we obtain Z Z Z Z Z K(x − y)(θ(y) − µ) dy dx + µ dx K(x − y)θ(y) dy dx = θ(x)>µ θ(x)>µ Rd θ(x)>µ Rd Z Z Z ≤ (θ(x) − µ) dx + µ dx = θ(x) dx. θ(x)>µ

θ(x)>µ

θ(x)>µ

ineg.rara

This completes the proof of (8.24).

¤ nonlocal.2

Control of the L1 -norm. As in the previous chapter, we multiply equation (8.2) by sgn(u(x, t)) and integrate in Rd to obtain the following estimate Z Z Z Z d |u(x, t)| dx = J(x − y)u(y, t)sgn(u(x, t)) dy dx − |u(t, x)| dx dt Rd Rd Rd Rd Z Z Z + G(x − y)f (u(y, t))sgn(u(x, t)) dy dx − f (u(x, t))sgn(u(x, t)) dx Rd Rd Rd Z Z Z ≤ J(x − y)|u(y, t)| dy dx − |u(x, t)| dx Rd Rd Rd Z Z Z + G(x − y)|f (u)(y, t)| dy dx − |f (u)(x, t)| dx Rd Rd Rd Z Z Z = |u(y, t)| J(x − y) dx dy − |u(x, t)| dx Rd Rd Rd Z Z Z + |f (u)(y, t)| G(x − y) dx dy − |f (u)(x, t)| dx Rd

Rd

Rd

≤ 0, which shows that the L1 -norm does not increase. Control of the L∞ -norm. Let us denote m = ku0 kL∞ (Rd ) . Multiplying the equation nonlocal.2 in (8.2) by sgn(u − m)+ and integrating in the x variable we get, Z d (u(x, t) − m)+ dx = I1 (t) + I2 (t) dt Rd where

Z

Z

Z +

I1 (t) =

u(x, t)sgn(u(x, t) − m)+ dx

J(x − y)u(y, t)sgn(u(x, t) − m) dy dx − Rd

Rd

Rd

96


and

Z

Z

I2 (t) = Rd

G(x − y)f (u)(y, t)sgn(u(x, t) − m)+ dy dx Rd Z − f (u)(x, t)sgn(u(x, t) − m)+ dx. Rd

We claim that both I1 and I2 are negative. Thus (u(x, t) − m)+ = 0 a.e. x ∈ Rd and then u(x, t) ≤ m for all t > 0 and a.e. x ∈ Rd . lema.ineg

In the case of I1 , applying Lemma 77 with K = J, θ = u(t) and µ = m we obtain Z Z Z Z + J(x − y)u(y, t) dy dx J(x − y)u(y, t)sgn(u(x, t) − m) dy dx = Rd Rd u(x)>m Rd Z ≤ u(x, t) dx. u(x)>m

lema.ineg

To handle the second one, I2 , we proceed in a similar manner. Applying Lemma 77 with θ(x) = f (u)(x, t) and µ = f (m) we obtain

Z

Z

Z G(x − y)f (u)(y, t) dy dx ≤

f (u(x,t))>f (m)

Rd

f (u)(x, t) dx. f (u(x,t))>f (m)

Using that f is a nondecreasing function, we rewrite this inequality in an equivalent form te obtain the desired inequality: Z Z G(x − y)f (u)(y, t)sgn(u(x, t) − m)+ dy dx Rd Rd Z Z = G(x − y)f (u)(y, t) dy dx u(x,t)≥m Rd Z Z = G(x − y)f (u)(y, t) dy dx f (u)(x,t)≥f (m) Rd Z ≤ f (u)(x, t) dx. u(x,t)≥m

ineg.rara.neg

In a similar way, by using inequality (8.25) we get Z d (u(x, t) + m)− dx ≤ 0, dt Rd which implies that u(x, t) ≥ −m for all t > 0 and a.e. x ∈ Rd . We conclude that ku(t)kL∞ (Rd ) ≤ ku0 kL∞ (Rd ) .


97

Step III. Uniqueness and contraction property. Let us consider u and v two solutions corresponding to initial data u0 and v0 respectively. We will prove that for any t > 0 the following holds: Z d |u(x, t) − v(x, t)| dx ≤ 0. dt Rd To this end, we multiply by sgn(u(x, t) − v(x, t)) the equation satisfied by u − v and using the symmetry of J, the positivity of J and G and that their mass equals one we obtain, Z Z Z d J(x − y)(u(y, t) − v(y, t))sgn(u(x, t) − v(x, t)) dx dy |u(x, t) − v(x, t)| dx = dt Rd Rd Rd Z Z |u(x, t) − v(x, t)| dx − Rd Rd Z Z + G(x − y)(f (u)(y, t) − f (v)(y, t))sgn(u(x, t) − v(x, t)) dx dy Rd Rd Z − |f (u)(x, t) − f (v)(x, t)| dx Rd Z Z Z ≤ J(x − y)|u(y, t) − v(y, t)| dx dy − |u(x, t) − v(x, t)| dx Rd Rd Rd Z Z Z + G(x − y)|f (u)(y, t) − f (v)(y, t)| dx dy − |f (u)(x, t) − f (v)(x, t)| dx Rd

Rd

Rd

= 0. Thus we get the uniqueness of the solutions and the contraction property ku(t) − v(t)kL1 (Rd ) ≤ ku0 − v0 kL1 (Rd ) . teo.existence

This ends the proof of Theorem 66.

¤

Now we prove that, due to the lack of regularizing effect, the L∞ (R)-norm does not 1 get bounded for positive times when we consider initial conditions in LEZ (R). This is in contrast to what happens for the local convection-diffusion problem, see [54]. unif.norm1

Proposition 78. Let d = 1 and |f (u)| ≤ C|u|q with 1 ≤ q < 2. Then 1

t 2 ku(t)kL∞ (R) sup sup = ∞. ku0 kL1 (R) u0 ∈L1 (R) t∈[0,1] Proof. Assume by contradiction that 1

river

(8.26)

t 2 ku(t)kL∞ (R) sup sup = M < ∞. ku0 kL1 (R) u0 ∈L1 (R) t∈[0,1]

ma.conver.1

adas.lineal

heat.lineal

98


integral.form

Using the representation formula (8.23) we get: °Z 1 ° ° ° ku(1)kL∞ (R) ≥ kS(1) ∗ u0 kL∞ (R) − ° S(1 − s) ∗ (G ∗ (f (u)) − f (u))(s) ds ° ° ° 0

L∞ (R)

teo.term.deriv

Using Lemma 75 the last term can be bounded as follows: Z 1 ° °Z 1 1 ° ° ≤ S(1 − s) ∗ (G∗(f (u)) − f (u))(s) ds ° h1 − si− 2 kf (u(s))kL∞ (R) ds ° ∞ L (R) 0 0 Z 1 Z 1 q q q q ≤C ku(s)kL∞ (R) ds ≤ CM ku0 kL1 (R) s− 2 ds 0

≤ CM q ku0 kqL1 (R) ,

0

provided that q < 2. This implies that the L∞ (R)-norm of the solution at time t = 1 satisfies ku(1)kL∞ (R) ≥ kS(1) ∗ u0 kL∞ (R) − CM q ku0 kqL1 (R) ≥ e−1 ku0 kL∞ (R) − kK1 kL∞ (R) ku0 kL1 (R) − CM q ku0 kqL1 (R) ≥ e−1 ku0 kL∞ (R) − Cku0 kL1 (R) − CM q ku0 kqL1 (R) . Choosing now a sequence u0,ε with ku0,ε kL1 (R) = 1 and ku0,ε kL∞ (R) → ∞ we obtain that ku0,ε (1)kL∞ (R) → ∞, river

a contradiction with our assumption (8.26). The proof of the result is now completed. ¤ 0.20. Convergence to the local problem. In this chapter we prove the convergence of solutions of the nonlocal problem to solutions of the local convection-diffusion equation when we rescale the kernels and let the scaling parameter go to zero. As we did in the previous sections we begin with the analysis of the linear part. Lemma 79. Assume that u0 ∈ L2 (Rd ). Let wε be the solution to Z   (w ) (x, t) = 1 Jε (x − y)(wε (y, t) − wε (x, t)) dy, ε t ε2 Rd (8.27)  wε (0, x) = u0 (x), and w the solution to (8.28)

(

wt (x, t) = ∆w(x, t), w(0, x) = u0 (x).

Then, for any positive T , lim sup kwε − wkL2 (Rd ) = 0.

ε→0 t∈[0,T ]


99

Rescaladas.lineal

Proof. Taking the Fourier transform in (8.27) we get ´ 1 ³b wε (t, ξ) − w cε (t, ξ) . w cε (t, ξ) = 2 Jε (ξ)c ε Therefore,

Ã

Jbε (ξ) − 1 w cε (t, ξ) = exp t ε2

! ub0 (ξ).

But we have, b Jbε (ξ) = J(εξ). Hence we get

Ã

b J(εξ) −1 w cε (t, ξ) = exp t ε2

! ub0 (ξ). heat.lineal

By Plancherel’s identity, using the well known formula for solutions to (8.28), 2

w(t, b ξ) = e−tξ ub0 (ξ). we obtain that Z kwε (t) −

w(t)k2L2 (Rd )

cond1

= R

¯ ¯2 b ¯ t J(εξ)−1 2¯ −tξ ¯e ε2 − e ¯ |ub0 (ξ)|2 dξ ¯ ¯ d

cond2

With R and δ as in (8.13) and (8.14) we get ¯ ¯2 Z Z b ¯ t J(εξ)−1 ¯ 2 ¯e ε2 − e−tξ2 ¯ |b ¯ ¯ u0 (ξ)| dξ ≤ |ξ|≥R/ε

lim.2

(8.29)

tδ

(e− ε2 + e

−tR2 ε2

)2 |ub0 (ξ)|2 dξ

|ξ|≥R/ε tδ

≤ (e− ε2 + e

−tR2 ε2

)2 ku0 k2L2 (Rd ) → 0. ε→0

To treat the integral on the set {ξ ∈ Rd : |ξ| ≤ R/ε} we use the fact that on this set the following holds:

est.100

(8.30)

¯ J(εξ)−1 ¯ ¯ J(εξ) ¯ b J(εξ)−1 −1 2 ¯ t b ε2 ¯b t −tξ 2 ¯ 2¯ ε2 −e + ξ max{e , e−tξ } ¯e ¯ ≤ t¯ ¯ 2 ε ¯ J(εξ) ¯ 2 b −1 2 ¯ 2¯ − tξ2 ≤ t¯ + ξ max{e , e−tξ } ¯ 2 ε ¯ J(εξ) ¯ 2 b −1 ¯ 2 ¯ − tξ2 ≤ t¯ + ξ e . ¯ ε2

100


Thus: Z |ξ|≤R/ε

liviu

¯ ¯2 Z b ¯ t J(εξ)−1 2¯ −tξ 2 ¯e ε2 − e ¯ |ub0 (ξ)| dξ ≤ ¯ ¯

¯ J(εξ) ¯2 −1 2 ¯b 2¯ e−t|ξ| t2 ¯ + ξ ¯ |ub0 (ξ)|2 dξ 2 ε |ξ|≤R/ε ¯ ¯2 Z ¯ J(εξ) 2 2¯ b − 1 + ε ξ 2 ¯ ¯ ≤ e−tξ t2 |ξ|4 ¯ ¯ |ub0 (ξ)|2 dξ. 2ξ2 ¯ ¯ ε |ξ|≤R/ε

b − 1| ≤ K|ξ|2 for all ξ ∈ Rd we get From |J(ξ) ¯ ¯ ¯ J(εξ) 2 2¯ b − 1 + ε ξ ¯ (K + 1) 2 2 ¯ (8.31) ¯ ¯ ≤ 2 2 ε |ξ| ≤ K + 1. ¯ ¯ ε2 ξ 2 ε |ξ| Using this bound and that e−|s| s2 ≤ C, we get that ¯2 ¯2 ¯ Z Z ¯¯ b 2 2¯ b ¯ t J(εξ)−1 ¯ J(εξ) − 1 + ε |ξ| 2 ¯ ¯ ¯e ε2 − e−tξ ¯ |ub0 (ξ)|2 dξ ≤ C sup u0 (ξ)|2 1{|ξ|≤R/ε} dξ. ¯ ¯ |b ¯ ¯ 2 2 ¯ ¯ ε |ξ| d t∈[0,T ] |ξ|≤R/ε R liviu

By inequality (8.31) together with the fact that b J(εξ) − 1 + ε2 |ξ|2 =0 ε→0 ε2 |ξ|2 lim

lim.1

.varepsilon

and that u b0 ∈ L2 (Rd ), by Lebesgue Z (8.32) lim sup ε→0 t∈[0,T ]

lim.2

dominated convergence theorem, we have that ¯ ¯2 b ¯ t J(εξ)−1 2¯ −tξ ¯e ε2 − e ¯ |ub0 (ξ)|2 dξ = 0. ¯ ¯

|ξ|≤R/ε

lim.1

From (8.29) and (8.32) we obtain lim sup kwε (t) − w(t)k2L2 (Rd ) = 0,

ε→0 t∈[0,T ]


¤

Next we prove a lemma that provides us with a uniform (independent of ε) decay for the nonlocal convective part. Lemma 80. There exists a positive constant C = C(J, G) such that °³ S (t) ∗ G − S (t) ´ ° 1 ° ε ° ε ε ∗ ϕ ° ° 2 d ≤ C t− 2 kϕkL2 (Rd ) ε L (R ) holds for all t > 0 and ϕ ∈ L2 (Rd ), uniformly on ε > 0. Here Sε (t) is the linear semigroup Rescaladas.lineal associated to (8.27).

en.gradient


101

Proof. Let us denote by Φε (x, t) the following quantity: Φε (x, t) =

(Sε (t) ∗ Gε )(x) − Sε (t)(x) . ε

Then by the definition of Sε and Gε we obtain Z ³ t(J(εξ) b b − 1) ´ G(ξε) −1 ix·ξ e exp Φε (x, t) = dξ 2 ε ε Rd Z ³ t(J(ξ) . b − 1) ´ iε−1 x·ξ −d−1 b − 1) dξ e exp =ε ( G(ξ) ε2 Rd −d−1 −2 −1 =ε Φ1 (tε , xε ) teo.term.deriv

At this point, we observe that for ε = 1, Lemma 75 gives us 1

kΦ1 (t) ∗ ϕkL2 (Rd ) ≤ C(J, G)hti− 2 kϕkL2 (Rd ) . Hence kΦε (t) ∗ ϕkL2 (Rd ) = ε−d−1 kΦ1 (tε−2 , ε−1 ·) ∗ ϕkL2 (Rd ) = ε−1 k[Φ1 (tε−2 ) ∗ ϕ(ε·)](ε−1 ·)kL2 (Rd ) d

d

1

= ε−1+ 2 kΦ1 (tε−2 ∗ ϕ(ε·))kL2 (Rd ) ≤ ε−1+ 2 (tε−2 )− 2 kϕ(ε·)kL2 (Rd ) 1

≤ t− 2 kϕkL2 (Rd ) . This ends the proof.

¤

Lemma 81. Let be T > 0 and M > 0. Then the following ° ¶ Z t °µ ° Sε (s) ∗ Gε − Sε (s) ° ° ° lim sup − b · ∇H(s) ∗ ϕ(s) ° 2 d ds = 0, ε→0 t∈[0,T ] 0 ° ε L (R ) holds uniformly for all kϕkL∞ ([0,T ];L2 (Rd )) ≤ M . Here H is the linear heat semigroup given by the Gaussian x2

H(t) = and b = (b1 , ..., bd ) is given by

e− 4t

d

(2πt) 2

Z

bj =

xj G(x) dx,

j = 1, ..., d.

Rd

Proof. Let us denote by Iε (t) the following quantity: ° ¶ Z t °µ ° Sε (s) ∗ Gε − Sε (s) ° ° ° Iε (t) = − b · ∇H(s) ∗ ϕ(s) ° ° 2 d ds. ε 0 L (R )

102


Choose α ∈ (0, 1). Then

( Iε (t) ≤

where

Z

if t ≤ εα ,

I1,ε + I2,ε (t)

if t ≥ εα ,

°µ ° ¶ ° Sε (s) ∗ Gε − Sε (s) ° ° − b · ∇H(s) ∗ ϕ(s)° ° ° ε

εα

I1,ε = 0

and

I1,ε

ds L2 (Rd )

° ¶ Z t °µ ° ° Sε (s) ∗ Gε − Sε (s) ° ° I2,ε (t) = − b · ∇H(s) ∗ ϕ(s) ° 2 d ds. ° ε α ε L (R )

The first term I1,ε satisfies, ° Z εα °³ Z εα ° Sε (s) ∗ Gε − Sε (s) ´ ° ° ° ds + kb · ∇H(s) ∗ ϕkL2 (Rd ) ds I1,ε ≤ ∗ ϕ° ° ε 0 0 L2 (Rd ) Z

s

≤ C 0

ineg.1

(8.33)

Z

εα

− 12

εα

k∇H(s)kL1 (Rd ) kϕ(s)kL2 (Rd ) ds

kϕ(s)kL2 (Rd ) ds + C 0

Z

εα

≤ CM

α

1

s− 2 ds = 2CM ε 2 .

0

To bound I2,ε (t) Z I2,ε (t) =

we observe that, by Plancherel’s identity, we get, °Ã ° ! ! Ã ° t ° b G(εξ) − 1 2 2 b ° s(J(εξ)−1)/ε ° − i b · ξe−s|ξ| ϕ(s) b ° ds ° e ° ° 2 d ε α ε Lξ (R ) ° ! Ã Z t° ° °³ ´ G(εξ) b −1 2 b ° ° s(J(εξ)−1)/ε −s|ξ|2 ϕ(s) b ≤ ds e − e ° ° ° 2 d ° ε εα Lξ (R ) ° ° Ã ! Z t° ° b −1 ° −s|ξ|2 G(εξ) ° + − i b · ξ ϕ(s) b ° ds °e ° 2 d ε εα ° Lξ (R ) Z t Z t = R1,ε (s) ds + R2,ε (s) ds. εα

εα

In the following we obtain upper bounds for R1,ε and R2,ε . Observe that R1,ε satisfies: (R1,ε )2 (s) ≤ 2((R3,ε )2 (s) + (R4,ε )2 (s)) where

¯ ¯2 ¯ ³ ´2 ¯ G(εξ) b − 1 2 2 b ¯ ¯ b ξ)|2 dξ (R3,ε )2 (s) = es(J(εξ)−1)/ε − e−s|ξ| ¯ ¯ |ϕ(s, ¯ ¯ ε |ξ|≤R/ε Z


103

and Z

³

(R4,ε )2 (s) =

2 b s(J(εξ)−1)/ε

e

−s|ξ|2

−e

|ξ|≥R/ε

¯ ¯2 ¯ ´2 ¯ G(εξ) b − 1 ¯ ¯ b ξ)|2 dξ. ¯ ¯ |ϕ(s, ¯ ¯ ε est.varepsilon

With respect to R3,ε we proceed as in the proof of Lemma 80 by choosing δ and R as cond1 cond2 est.100 b in (8.13) and (8.14). Using estimate (8.30) and the fact that |G(ξ) − 1| ≤ C|ξ| and 2 3 d b |J(ξ) − 1 + ξ | ≤ C|ξ| for every ξ ∈ R we obtain: ¯2 ¯ ¯ J(εξ) 2 2¯ b − 1 + ξ ε ¯ ¯ b ξ)|2 dξ C e s2 ¯ ¯ |ξ|2 |ϕ(s, 2 ¯ ¯ ε |ξ|≤R/ε · ¸ Z 3 2 −s|ξ|2 2 (εξ) e s C |ξ|2 |ϕ(s, b ξ)|2 dξ 2 ε |ξ|≤R/ε Z Z 2 −s|ξ|2 2 2 8 2 2 −2 C e s ε |ξ| |ϕ(s, b ξ)| dξ ≤ ε s e−s|ξ| s4 |ξ|8 |ϕ(s, b ξ)|2 dξ d |ξ|≤R/ε R Z Cε2−2α |ϕ(s, b ξ)|2 dξ ≤ Cε2−2α M 2 . Z

(R3,ε )2 (s) ≤ ≤ = ≤

−s|ξ|2

Rd

lim.2 ˆ In the case of R4,ε , we use that |G(ξ)| ≤ 1 and we proceed as in the proof of (8.29):

Z sδ

(e− ε2 + e−

2

(R4,ε ) (s) ≤ |ξ|≥R/ε δ

≤ (e− ε2−α

sR2 ε2

)2 ε−2 |ϕ(s, b ξ)|2 dξ Z R2 − 2−α 2 −2 +e ε ) ε |ϕ(s, b ξ)|2 dξ |ξ|≥R/ε

δ − 2−α ε

R2 − 2−α ε

≤ M 2 (e +e 2 2−2α ≤ CM ε

)2 ε−2

for sufficiently small ε. Then Z ineg.2

t

(8.34) εα

R1,ε (s)ds ≤ CT M ε1−α .

104


b − 1 − ib · ξ| ≤ C|ξ|2 The second term can be estimated in a similar way, using that |G(ξ) for every ξ ∈ Rd , we get ¯ ¯2 Z ¯ G(εξ) ¯ b − 1 − i b · ξε 2 ¯ ¯ 2 −2s|ξ| (R2,ε ) (s) ≤ e b ξ)|2 dξ ¯ ¯ |ϕ(s, ¯ ¯ ε Rd · ¸ Z Z 2 2 2 −2s|ξ|2 (ξε) 2 ≤ C e |ϕ(s, b ξ)| dξ = C e−2s|ξ| ε2 |ξ|4 |ϕ(s, b ξ)|2 dξ ε Rd Rd Z Z −2s|ξ|2 2 4 2 2(1−α) 2 −2 |ϕ(s, b ξ)|2 dξ e s |ξ| |ϕ(s, b ξ)| dξ ≤ Cε = Cε s Rd 2 2(1−α)

≤ CM ε and we conclude that ineg.3

Rd

, Z

t

(8.35)

R2,ε (s)ds ≤ CT M ε1−α .

εα

ineg.1 ineg.2

ineg.3

Now, by (8.33), (8.34) and (8.35) we obtain that pepe.2

α

sup Iε (t) ≤ CM (ε 2 + ε1−α ) → 0,

(8.36)

as ε → 0,

t∈[0,T ]


¤ teo.convergencia.ve.to.zero

Now we are ready to prove Theorem 67.

teo.convergencia.ve.to.zero

Proof of Theorem 67. First we write the two problems in the semigroup formulation, Z t Sε (t − s) ∗ Gε − Sε (t − s) ∗ f (uε (s)) ds uε (t) = Sε (t) ∗ u0 + ε 0 and Z t v(t) = H(t) ∗ u0 + b · ∇H(t − s) ∗ f (v(s)) ds. 0

Then ineg.5

(8.37)

sup kuε (t) − v(t)kL2 (Rd ) ≤ sup I1,ε (t) + sup I2,ε (t)

t∈[0,T ]

t∈[0,T ]

t∈[0,T ]

where I1,ε (t) = kSε (t) ∗ u0 − H(t) ∗ u0 kL2 (Rd ) and

°Z ° I2,ε (t) = ° °

t 0

Sε (t − s) ∗ Gε − Sε (t − s) ∗ f (uε (s)) − ε lema.conver.1

Z

t 0

° ° b · ∇H(t − s) ∗ f (v(s))° °

In view of Lemma 79 we have

sup I1,ε (t) → 0 t∈[0,T ]

as ε → 0.

. L2 (Rd )


105

So it remains to analyze the second term I2,ε . To this end, we split it again I2,ε (t) ≤ I3,ε (t) + I4,ε (t) where

and

ineg.6

° Z t° ° Sε (t − s) ∗ Gε − Sε (t − s) ¡ ¢° ° I3,ε (t) = ∗ f (uε (s)) − f (v(s)) ° ° ° 2 d ds ε 0 L (R )

° ¶ Z t °µ ° Sε (t − s) ∗ Gε − Sε (t − s) ° ° ° I4,ε (t) = − b · ∇H(t − s) ∗ f (v(s)) ° ° 2 d ds. ε 0 L (R )

Using Young’s inequality and that from our hypotheses we have an uniform bound for uε and u in terms of ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) we obtain Z t kf (uε (s)) − f (v(s))kL2 (Rd ) I3,ε (t) ≤ ds 1 |t − s| 2 0 Z t ds (8.38) ≤ kf (uε ) − f (v)kL∞ ((0,T ); L2 (Rd )) 1 0 |t − s| 2 ≤ 2T 1/2 kuε − vkL∞ ((0,T ); L2 (Rd )) C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ). temen.gradient

pepe.2

By Lemma 81 we obtain, choosing α = 2/3 in (8.36), that ineg.7

1

(8.39)

1

sup I4,ε ≤ Cε 3 kf (v)kL∞ ((0,T ); L2 (Rd )) ≤ Cε 3 C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ).

t∈[0,T ]

ineg.5 ineg.6

ineg.7

Using (8.37), (8.38) and (8.39) we get: kuε − vkL∞ ((0,T ); L2 (Rd )) ≤ kI1,ε kL∞ ((0,T ); L2 (Rd )) 1

+T 2 C(ku0 kL1 (R) , ku0 kL∞ (R) )kuε − vkL∞ ((0,T ); L2 (Rd )) . Choosing T = T0 sufficiently small, depending on ku0 kL1 (R) and ku0 kL∞ (R) we get kuε − vkL∞ ((0,T ); L2 (Rd )) ≤ kI1,ε kL∞ ((0,T ); L2 (Rd )) → 0, as ε → 0. Using the same argument in any interval [τ, τ + T0 ], the stability of the solutions of the Rescaladas equation (8.4) in L2 (Rd )-norm and that for any time τ > 0 it holds that kuε (τ )kL1 (Rd ) + kuε (τ )kL∞ (Rd ) ≤ ku0 kL1 (Rd ) + ku0 kL∞ (Rd ) , we obtain lim sup kuε − vkL2 (Rd ) = 0,

ε→0 t∈[0,T ]


¤

106


0.21. Long time behaviour of the solutions. The aim ofnonlocal.2 this chapter is to obtain the first term in the asymptotic expansion of the solution u to (8.2). The main ingredient for our proofs is the following lemma inspired in the Fourier splitting method introduced MR571048 MR856876 MR1356749 by Schonbek, see [78], [79] and [80]. puto.lema hip.1

Lemma 82. Let R and δ be such that the function Jb satisfies: 2 b ≤ 1 − |ξ| , J(ξ) 2

(8.40)

|ξ| ≤ R

and hip.2

ineg.dif

puta.ineg

b ≤ 1 − δ, J(ξ)

(8.41)

|ξ| ≥ R.

Let us assume that the function u : [0, ∞) × Rd → R satisfies the following differential inequality: Z Z d 2 |u(x, t)| dx ≤ c (J ∗ u − u)(x, t)u(x, t) dx, (8.42) dt Rd Rd for any t > 0. Then for any 1 ≤ r < ∞ there exists a constant a = rd/cδ such that Z ku(0)k2L2 (Rd ) rdω0 (2δ) d2 Z t d 2 (8.43) + (s + 1)rd− 2 −1 ku(as)k2L1 (Rd ) ds |u(at, x)| dx ≤ rd rd (t + 1) (t + 1) Rd 0 holds for all positive time t where ω0 is the volume of the unit ball in Rd . In particular

ineg.final

(8.44)

ku(at)kL2 (Rd ) ≤

ku(0)kL2 (Rd ) rd

(t + 1) 2

1

+

d

(2ω0 ) 2 (2δ) 4 d

(t + 1) 4

kukL∞ ([0,∞); L1 (Rd )) .

hip.1 b ≤ 1 − A|ξ|2 for |ξ| ≤ R but Remark 83. Condition (8.40) can be replaced by J(ξ) omitting the constant A in the proof we simplify some formulas. ineg.dif

Remark 84. The differential inequality (8.42) can be written in the following form: Z Z Z d c 2 |u(x, t)| dx ≤ − J(x − y)(u(x, t) − u(y, t))2 dx dy. dt Rd 2 Rd Rd EZ

This is the nonlocal version of the energy method used in [54]. However, in our case, EZ exactly the same inequalities used in [54] could not be applied. hip.1

hip.2

Proof. Let R and δ be as in (8.40) and (8.41). We set a = rd/cδ and consider the following set: ( µ ¶1/2 ) 2rd . A(t) = ξ ∈ Rd : |ξ| ≤ M (t) = c(t + a) ineg.dif

ineg.difff

Inequality (8.42) gives us: Z Z Z d 2 2 b b − 1)|b (8.45) |u(x, t)| dx ≤ c (J(ξ) − 1)|b u(ξ)| dξ ≤ c (J(ξ) u(ξ)|2 dξ. dt Rd d c R A(t)


hip.1

ineg.j

107

hip.2

Using the hypotheses (8.40) and (8.41) on the function Jb the following inequality holds for all ξ ∈ A(t)c : b − 1) ≤ − rd , (8.46) c(J(ξ) for every ξ ∈ A(t)c , t+a since for any |ξ| ≥ R b − 1) ≤ −cδ = − rd ≤ − rd c(J(ξ) a t+a and 2 b − 1) ≤ − c|ξ| ≤ − c 2rd = − rd c(J(ξ) 2 2 c(t + a) t+a c for all ξ ∈ A(t) with |ξ| ≤ R. ineg.j

ineg.difff

Introducing (8.46) in (8.45) we obtain Z Z rd d 2 |u(x, t)| dx ≤ − |b u(t, ξ)|2 dξ dt Rd t + a A(t)c Z Z rd rd 2 ≤ − |b u(t, ξ)|2 dξ |b u(t, ξ)| dξ + t + a Rd t + a |ξ|≤M (t) Z rd rd ≤ − M (t)d ω0 kb u(t)k2L∞ (Rd ) |u(x, t)|2 dx + t + a Rd t+a · ¸ d2 Z 2rd rd rd ≤ − |u(x, t)|2 dx + ω0 ku(t)k2L1 (Rd )) . t + a Rd t + a c(t + a) This implies that Z i dh rd 2 (t + a) |u(x, t)| dx dt Rd · Z ¸ Z d rd 2 rd−1 = (t + a) |u(x, t)| dx + rd(t + a) |u(x, t)|2 dx dt Rd Rd µ ¶d 2rd 2 rd− d2 −1 rd ω0 ku(t)k2L1 (Rd ) . ≤ (t + a) c Integrating on the time variable the last inequality we obtain: µ ¶d Z Z Z d 2rd 2 t rd 2 rd 2 (t+a) |u(x, t)| dx−a |u(0, x)| dx ≤ rdω0 (s+a)rd− 2 −1 ku(s)k2L1 (Rd ) ds c Rd Rd 0 and hence Z Z ard 2 |u(x, t)| dx ≤ |u(0, x)|2 dx rd (t + a) d d R R µ ¶d Z d rdω0 2rd 2 t + (s + a)rd− 2 −1 ku(s)k2L1 (Rd ) ds. rd (t + a) c 0

108


Replacing t by ta we get: Z 2

|u(at, x)| dx ≤

ku(0)k2L2 (Rd ) (t + 1)rd

Rd

= =

ku(0)k2L2 (Rd ) (t + 1)rd ku(0)k2L2 (Rd )

puta.ineg

(t + 1)rd

µ ¶d Z d 2rd 2 at rdω0 + (s + a)rd− 2 −1 ku(s)k2L1 (Rd ) ds rd rd (t + 1) a c 0 µ ¶ d2 Z t d rdω0 2rd + (s + 1)rd− 2 −1 ku(as)k2L1 (Rd ) ds rd (t + 1) ca 0 d Z t d rdω0 (2δ) 2 + (s + 1)rd− 2 −1 ku(as)k2L1 (Rd ) ds rd (t + 1) 0

which proves (8.43).

ineg.final

Estimate (8.44) is obtained as follows: Z Z t ku(0)k2L2 (Rd ) rdω0 (2δ) d2 d 2 2 |u(at, x)| dx ≤ + kukL∞ ([0,∞); L1 (Rd )) (s + 1)rd− 2 −1 ds rd rd (t + 1) (t + 1) Rd 0 d 2 ku(0)kL2 (Rd ) 2ω0 (2δ) 2 2 ≤ + d kukL∞ ([0,∞); L1 (Rd )) . rd (t + 1) (t + 1) 2 This ends the proof. diff.ineq

¤

Lemma 85. Let 2 ≤ p < ∞. For any function u : Rd 7→ Rd , I(u) defined by Z I(u) = (J ∗ u − u)(x)|u(x)|p−1 sgn(u(x)) dx Rd

satisfies

Z 4(p − 1) I(u) ≤ (J ∗ |u|p/2 − |u|p/2 )(x)|u(x)|p/2 dx 2 p Rd Z Z 2(p − 1) =− J(x − y)(|u(y)|p/2 − |u(x)|p/2 )2 dx dy. p2 Rd Rd

Remark 86. This result is a nonlocal counterpart of the well known identity Z Z 4(p − 1) p−1 ∆u |u| sgn(u) dx = − |∇(|u|p/2 )|2 dx. p2 Rd Rd Proof. Using the symmetry of J, I(u) can be written in the following manner, Z Z I(u) = J(x − y)(u(y) − u(x))|u(x)|p−1 sgn(u(x)) dx dy d d ZR ZR = J(x − y)(u(x) − u(y))|u(y)|p−1 sgn(u(y)) dx dy. Rd

Thus 1 I(u) = − 2

Z

Z Rd

Rd

¡ ¢ J(x − y)(u(x) − u(y)) |u(x)|p−1 sgn(u(x)) − |u(y)|p−1 sgn(u(y)) dx dy. Rd


109

Using the following inequality, ||α|p/2 − |β|p/2 |2 ≤

p2 (α − β)(|α|p−1 sgn(α) − |β|p−1 sgn(β)) 4(p − 1)

which holds for all real numbers α and β and for every 2 ≤ p < ∞, we obtain that I(u) can be bounded from above as follows: Z Z 4(p − 1) I(u) ≤ − J(x − y)(|u(y)|p/2 − |u(x)|p/2 )2 dx dy 2p2 d d ZR ZR 4(p − 1) = − J(x − y)(|u(y)|p − 2|u(y)|p/2 |u(x)|p/2 + |u(x)|p ) dx dy 2p2 d d Z R R 4(p − 1) = (J ∗ |u|p/2 − |u|p/2 )(x)|u(x)|p/2 dx. p2 d R The proof is finished.

¤ long.time

Now we are ready to proceed with the proof of Theorem 69. long.time

Proof of Theorem 69. Let u be the solution to the nonlocal convection-diffusion problem. Then, by the same arguments that we used to control the L1 (Rd )-norm, we obtain the following: Z Z d p |u(x, t)| dx = p (J ∗ u − u)(x, t)|u(x, t)|p−1 sgn(u(x, t)) dx dt Rd d ZR + (G ∗ f (u) − f (u))(x, t)|u(x, t)|p−1 sgn(u(x, t)) dx d ZR ≤ p (J ∗ u − u)(x, t)|u(x, t)|p−1 sgn(u(x, t)) dx. Rd

diff.ineq

Using Lemma 85 we get that the Lp (Rd )-norm of the solution u satisfies the following differential inequality: ineg.100

(8.47)

d dt

Z

4(p − 1) |u(x, t)| dx ≤ p Rd

Z

p

(J ∗ |u|p/2 − |u|p/2 )(x)|u(x)|p/2 dx. Rd

First, let us consider p = 2. Then Z Z d 2 |u(x, t)| dx ≤ 2 (J ∗ |u| − |u|)(x, t)|u(x, t)| dx. dt Rd Rd

110


puto.lema

Applying Lemma 82 with |u|, c = 2, r = 1 and using that kukL∞ ([0,∞); L1 (Rd )) ≤ ku0 kL1 (Rd ) we obtain 1 d ku0 kL2 (Rd ) (2ω0 ) 2 (2δ) 4 ku(td/2δ)kL2 (R) ≤ + kukL∞ ([0,∞); L1 (Rd )) d d (t + 1) 2 (t + 1) 4 ku0 kL2 (Rd )

≤

decay.rate

1

d

(2ω0 ) 2 (2δ) 4

+ ku0 kL1 (Rd )) d d (t + 1) 2 (t + 1) 4 C(J, ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) , ≤ d (t + 1) 4

nonlocal.2

which proves (8.7) in the case p = 2. Using that the L1 (Rd )-norm of the solutions to (8.2), does not increase, ku(t)kL1 (Rd ) ≤ ku0 kL1 (Rd ) , by Hölder’s inequality we obtain the desired decay.rate decay rate (8.7) in any Lp (Rd )-norm with p ∈ [1, 2]. In the following, using an inductive argument, we will prove the result for any r = 2m , with m ≥ 1 an integer. By Hölder’s inequality this will give us the Lp (Rd )-norm decay for any 2 < p < ∞. Let us choose r = 2m with m ≥ 1 and assume that the following d

1

ku(t)kLr (Rd ) ≤ Chti− 2 (1− r ) holds for some positive constant C = C(J, ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) and for every positive time t. We want to show an analogous estimate for p = 2r = 2m+1 . ineg.100 We use (8.47) with p = 2r to obtain the following differential inequality: Z Z d 4(2r − 1) 2r |u(x, t)| dx ≤ (J ∗ |u|r − |u|r )(x, t)|u(x, t)|r dx. dt Rd 2r d R puto.lema

Applying Lemma (82) with |u|r , c(r) = 2(2r − 1)/r and a = rd/c(r)δ we get: Z kur0 k2L2 (Rd ) dω0 (2δ) d2 Z t d 2r |u(at)| ≤ + (s + 1)rd− 2 −1 kur (as)k2L1 (Rd ) ds rd rd (t + 1) (t + 1) Rd 0 Z t 2r ku0 kL2r (Rd ) d C(J) (s + 1)rd− 2 −1 ku(as)k2r ≤ + Lr (Rd ) ds rd rd (t + 1) (t + 1) 0 C(J, ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) ≤ × (t + 1)d µ ¶ Z t rd− d2 −1 −dr(1− r1 ) 1+ (s + 1) (s + 1) ds 0 d d C ≤ (1 + (t + 1) 2 ) ≤ C(t + 1) 2 −dr dr (t + 1)

and then

d

1

ku(at)kL2r (Rd ) ≤ C(J, ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) )(t + 1)− 2 (1− 2r ) ,


111


¤

Let us close this chapter with a remark concerning the applicability of energy methods to study nonlocal problems.

mark.energy

Remark 87. If we want to use energy estimates to get decay rates (for example in L2 (Rd )), we arrive easily to Z Z Z 1 d 2 |w(x, t)| dx = − J(x − y)(w(x, t) − w(y, t))2 dx dy dt Rd 2 Rd Rd when we deal with a solution of the linear equation wt = J ∗ w − w and to Z Z Z d 1 2 |u(x, t)| dx ≤ − J(x − y)(u(x, t) − u(y, t))2 dx dy dt Rd 2 Rd Rd when we consider the complete convection-diffusion problem. However, we can not go further since an inequality of the form µZ ¶ p2 Z Z p |u(x)| dx ≤C J(x − y)(u(x) − u(y))2 dx dy Rd

Rd

Rd

is not available for p > 2. 0.22. Weakly nonlinear behaviour. In this section we find the leading order term nonlocal.2 EZ in the asymptotic expansion of the solution to (8.2). We use ideas from [54] showing that the nonlinear term decays faster than the linear part. liv.rossi

We recall a previous result of [68] that extends to nonlocal diffusion problems the result DZ of [52] in the case of the heat equation. Lemma 88. Let J ∈ S(Rd ) such that b − (1 − |ξ|2 ) ∼ B|ξ|3 , J(ξ)

ξ ∼ 0,

for some constant B. For every p ∈ [2, ∞), there exists some positive constant C = C(p, J) such that decay.heat

d

1

1

kS(t) ∗ ϕ − M H(t)kLp (Rd ) ≤ Ce−t kϕkLp (Rd ) + CkϕkL1 (Rd ,|x|) hti− 2 (1− p )− 2 , t > 0, R for every ϕ ∈ L1 (Rd , 1 + |x|) with M = R ϕ(x) dx, where (8.48)

x2

H(t) =

e− 4t

d

(2πt) 2

,

is the gaussian. b − (1 − A|ξ|2 ) ∼ B|ξ|3 for ξ ∼ 0 Remark 89. We can consider a condition like J(ξ) and obtain as profile a modified Gaussian HA (t) = H(At), but we omit the tedious details.

112


Remark 90. The case p ∈ [1, 2) is more subtle. The analysis performed in the previous sections to handle the case p = 1 can be also extended to cover this case when the dimension b ∼ 1 − A|ξ|s , ξ ∼ 0, then s has d verifies 1 ≤ d ≤ 3. Indeed in this case, if J satisfies J(ξ) to be grater than [d/2] + 1 and s = 2 to obtain the Gaussian profile. Proof. We write S(t) = e−t δ0 + Kt . Then it is sufficient to prove that d

1

1

kKt ∗ ϕ − M Kt kLp (Rd ) ≤ CkϕkL1 (Rd ,|x|) hti− 2 (1− p )− 2 and d

1

1

t 2 (1− p ) kKt − H(t)kLp (Rd ) ≤ Chti− 2 . nucleu b A The first estimate follows by Lemma 74. The second one uses the hypotheses on J. liv.rossi detailed proof can be found in [68]. ¤

Now we are ready to prove that the same expansion holds for solutions to the complete nonlocal.2 problem (8.2) when q > (d + 1)/d. teo.weak.behavior decay.heat

Proof of Theorem 70. In view of (8.48) it is sufficient to prove that d

1

d

1

t− 2 (1− p ) ku(t) − S(t) ∗ u0 kLp (Rd ) ≤ Chti− 2 (q−1)+ 2 . integral.form

Using the representation (8.23) we get that Z

t

ku(t) − S(t) ∗ u0 kLp (Rd ) ≤

k[S(t − s) ∗ G − S(t − s)] ∗ |u(s)|q−1 u(s)kLp (Rd ) ds.

0

We now estimate the right hand side term as follows: we will split it in two parts, one in which we integrate on (0, t/2)teo.term.deriv and anotherlong.time one where we integrate on (t/2, t). Concerning the second term, by Lemma 75, Theorem 69 we have, Z t k[S(t − s) ∗ G − S(t − s)] ∗ |u(s)|q−1 u(s)kLp (Rd ) ds t/2

Z

t

≤ C(J, G) t/2

1

ht − si− 2 ku(s)kqLpq (Rd ) ds Z

t

≤ C(J, G, ku0 kL1 (Rd ) , ku0 kL∞ (R) )

1

t/2

≤ Chti

− d2 (q− p1 )+ 12

≤ Ct

− d2 (1− p1 )

d

d

1

ht − si− 2 hsi− 2 (q− p ) ds 1

hti− 2 (q−1)+ 2 .


113

To bound the first term we proceed as follows, Z t/2 k[S(t − s) ∗ G − S(t − s)] ∗ |u(s)|q−1 u(s)kLp (Rd ) ds 0

Z

t/2

≤ C(p, J, G) 0

≤ Chti

− d2 (1− p1 )− 12

d

³Z

t/2 0

= Chti

− d2 (1− p1 )− 12

1

1

ht − si− 2 (1− p )− 2 (k|u(s)|q kL1 (Rd ) + k|u(s)|q kLp (Rd ) ) ds Z ku(s)kqLq (Rd ) ds

t/2

+ 0

ku(s)kqLpq (Rd ) ds

´

(I1 (t) + I2 (t)).

long.time

By Theorem 69, for the first integral, I1 (t), we have the following estimate: Z t/2 Z t/2 d q I1 (t) ≤ ku(s)kLq (Rd ) ds ≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) ) hsi− 2 (q−1) ds, 0

0

and an explicit computation of the last integral shows that Z t/2 d d 1 − 12 hti hsi− 2 (q−1) ds ≤ Chti− 2 (q−1)+ 2 . 0

Arguing in the same manner for I2 we get hti

− 12

I2 (t) ≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) )hti ≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) )hti

Z − 12

t/2

dq

1

hsi− 2 (1− pq ) ds

0 − d2 (q− p1 )+ 12 d

1

≤ C(ku0 kL1 (Rd ) , ku0 kL∞ (Rd ) )hti− 2 (q−1)+ 2 . This ends the proof.

¤

CHAPTER 9

A nonlinear Neumann problem

.nonlin.Neu

defdweakIII

In this chapter we turn our attention to nonlinear equations with Neumann boundary conditions. We study Z    z (t, x) = J(x − y)(u(t, y) − u(t, x)) dy, x ∈ Ω, t > 0,   t Ω PγJ (z0 ) z(t, x) ∈ γ(u(t, x)), x ∈ Ω, t > 0,     z(0, x) = z0 (x), x ∈ Ω. Here Ω is a bounded domain, z0 ∈ L1 (Ω) and γ is a maximal monotone graph in R2 such that 0 ∈ γ(0). Solutions to PγJ (z0 ) will be understood in the following sense. Definition 91. A solution of PγJ (z0 ) in [0, T ] is a function z ∈ W 1,1 (]0, T [; L1 (Ω)) which satisfies z(0, x) = z0 (x), a.e. x ∈ Ω, and for which there exists u ∈ L2 (0, T ; L2 (Ω)), z ∈ γ(u) a.e. in QT = Ω×]0, T [, such that Z zt (t, x) = J(x − y)(u(t, y) − u(t, x)) dy a.e in ]0, T [×Ω. Ω

The main results of this chapter can be summarized as follows. “Under some natural assumptions about the initial condition z0 , there exists a unique global solution to PγJ (z0 ). Moreover, a contraction principle holds, given two solutions zi of PγJ (zi 0 ), i = 1, 2, then Z Z + (z1 (t) − z2 (t)) ≤ (z1 0 − z2 0 )+ . Ω

Ω

Respect to the asymptotic behaviour of the solution we prove that if γ is a continuous function, then Z 1 z0 , lim z(t) = t→∞ |Ω| Ω strongly in L1 (Ω)”. We can consider different maximal monotone graphs γ. For example, if γ(r) = rm , problem PγJ (z0 ) corresponds to the nonlocal version of the porous medium (or fast diffusion) 115

116

9. A NONLINEAR NEUMANN PROBLEM

problems. Note also that γ may be multivalued, so we are considering the nonlocal version of various phenomena with phase changes like the multiphase Stefan problem, for which  if r < 0,  r−1 [−1, 0] if r = 0, γ(r) =  r if r > 0. Even γ can have a domain different from R, which corresponds to obstacle problems. 0.23. Notations and preliminaries. In this section we collect some preliminaries and notations that will be used in the sequel. For a maximal monotone graph η in R × R and r ∈ N we denote by ηr the Yosida approximation of η, given by ηr = r(I − (I + 1r η)−1 ). The function ηr is maximal monotone and Lipschitz. We recall the definition of the main section η 0 of η    the element of minimal absolute value of η(s) if η(s) 6= ∅, 0 +∞ if [s, +∞) ∩ D(η) = ∅, η (s) :=   −∞ if (−∞, s] ∩ D(η) = ∅, where D(η) denotes the domain of η. The following properties hold: if s ∈ D(η), |ηr (s)| ≤ |η 0 (s)| and ηr (s) → η 0 (s) as r → +∞, and if s ∈ / D(η), |ηr (s)| → +∞ as r → +∞. We will use the following notations, η− := inf Ran(η) R r and η+ := sup Ran(η), where Ran(η) denotes the range of η. If 0 ∈ D(η), jη (r) = 0 η 0 (s)ds defines a convex l.s.c. function such that η = ∂jη . If jη∗ is the Legendre transform of jη then η −1 = ∂jη∗ . Also we will denote by J0 and P0 the following sets of functions, J0 = {j : R → [0, +∞], convex and lower semi-continuos with j(0) = 0}, BC

P0 = {p ∈ C ∞ (R) : 0 ≤ p0 ≤ 1, supp(p0 ) is compact, and 0 ∈ / supp(p)} .

In [19] the following relation for u, v ∈ L1 (Ω) is defined, Z Z u ¿ v if and only if j(u) dx ≤ j(v) dx, Ω

Ω

and the following facts are proved. 00II

Proposition 92. Let Ω be a bounded domain in RN . R R (i) For any u, v ∈ L1 (Ω), if Ω up(u) ≤ Ω vp(u) for all p ∈ P0 , then u ¿ v. (ii) If u, v ∈ L1 (Ω) and u ¿ v, then kukq ≤ kvkq for any q ∈ [1, +∞]. (iii) If v ∈ L1 (Ω), then {u ∈ L1 (Ω) : u ¿ v} is a weakly compact subset of L1 (Ω). ChChR

sec.lineal

The following Poincaré’s type inequality is given in [34], see also Chapter 2.


blg:01.II

PoincarejII

117

Proposition 93. Given J and Ω the quantity Z Z 1 J(x − y)(u(y) − u(x))2 dy dx 2 Ω Ω Z (9.1) β1 := β1 (J, Ω) = infR 2 u∈L (Ω), Ω u=0 (u(x))2 dx Ω

Poincare.II

is strictly positive. Consequently Z ¯ Z ¯2 Z Z ¯ ¯ 1 1 ¯ ¯ (9.2) β1 ¯u − u¯ ≤ J(x − y)(u(y) − u(x))2 dy dx, |Ω| 2 Ω Ω Ω Ω

∀u ∈ L2 (Ω).

In order to obtain a generalized Poincaré’s type inequality we need the following result. teo:6:2

ecua:p02

ecua:p01

Proposition 94. Let Ω ⊂ RN be a bounded open set and k > 0. There exists a constant C > 0 such that for any K ⊂ Ω with |K| > k, it holds Ã° ¯Z ¯! Z ° ° ° ¯ ¯ 1 ¯ ¯ (9.3) kukL2 (Ω) ≤ C ° u° ∀ u ∈ L2 (Ω). °u − |Ω| ° 2 + ¯ u¯ , Ω K L (Ω) Proof. Assume the conclusion does not hold. Then, for every n ∈ N there exists Kn ⊂ Ω with |Kn | > k, and un ∈ L2 (Ω) satisfying Ã° ° ¯Z ¯! Z ° ° ¯ ¯ 1 ¯ (9.4) kun kL2 (Ω) ≥ n ° un ° un ¯¯ , ∀ n ∈ N. °un − |Ω| ° 2 +¯ Ω Kn L (Ω) We normalize un by kun kL2 (Ω) = 1 for all n ∈ N, and consequently we can assume that

=ecua:p04

(9.5)

un * u

ecua:p01

ecua:p04

Moreover, by (9.4), we have ° ° Z ° ° 1 ° un − ° (9.6) u n ° ° |Ω| Ω

L2 (Ω)

1 ≤ , n

weakly in L2 (Ω).

and

¯Z ¯ ¯ ¯

¯ ¯ 1 un ¯¯ ≤ , n Kn

∀ n ∈ N.

Hence

Z 1 un − un → 0 in L2 (Ω), |Ω| Ω R =ecua:p04 1 and by (9.5) we get u(x) = |Ω| u = α for almost all x ∈ Ω, and un → α strongly in Ω ecua:p04 2 L (Ω). Since kun kL2 (Ω) = 1 for each n ∈ N, α 6= 0. On the other hand, (9.6) implies Z lim un = 0. n→∞

Kn

Since χKn is bounded in L2 (Ω), we can extract a subsequence (still denoted by χKn ) such that χKn * φ weakly in L2 (Ω).

118


Moreover, φ is nonnegative and verifies

Z

Z χKn =

k ≤ lim |Kn | = lim n→∞

n→∞

Ω

φ. Ω

Now, since un → α strongly in L2 (Ω) and χKn → φ weakly in L2 (Ω) we have Z Z Z χKn un = α φ 6= 0, 0 = lim un = lim n→∞

n→∞

Kn

Ω

Ω

a contradiction.

¤

To simplify the notation we define the linear self-adjoint operator A : L2 (Ω) → L2 (Ω) by

Z Au(x) =

J(x − y)(u(y) − u(x)) dy,

x ∈ Ω.

Ω

As a consequence of the above results we have the next proposition, which plays the role of the classical generalized Poincaré’s inequality for Sobolev spaces. teo:6

ecua:p05

Proposition 95. Let Ω ⊂ RN be a bounded open set and k > 0. There exists a constant C = C(J, Ω, k) such that, for any K ⊂ Ω with |K| > k, Ãµ Z ! ¶1/2 (9.7) kukL2 (Ω) ≤ C − Au u + kukL2 (K) ∀u ∈ L2 (Ω). Ω

AIMTp

Using the above result and working as in the proof of Lemma 4.2 in [4], we obtain the following lemma, of which we give the proof for the sake of completeness. lemma:01

Lemma 96. Let γ be a maximal monotone graph in R2 such that 0 ∈ γ(0). Let {un }n∈N ⊂ L2 (Ω) and {zn }n∈N ⊂ L1 (Ω) such that, for every n ∈ N, zn ∈ γ(un ) a.e. in Ω. Let us suppose that (i) if γ+ = +∞, there exists M > 0 such that Z zn+ < M,

∀n ∈ N,

Ω

(ii) if γ+ < +∞, there exists M ∈ R and h > 0 such that Z zn < M < γ+ |Ω|, ∀n ∈ N Ω

and

Z |zn | < {x∈Ω:zn (x) 0 such that Z zn− < M,

∀n ∈ N,

Ω

(iv) if γ− > −∞, there exists M ∈ R and h > 0 such that Z zn > M > γ− |Ω|, ∀n ∈ N Ω

and

Z zn < {x∈Ω:zn (x)>h}

her02

M − γ− |Ω| , 4

∀n ∈ N.

˜ ˜ Then, there exists a constant C˜ = C(M, Ω) in case (iii), C˜ = C(M, Ω, γ, h) in case (iv), such that Ãµ Z ! ¶1/2 − ˜ (9.9) ku− − Au− +1 , ∀n ∈ N. n kL2 (Ω) ≤ C n un Ω

her01

her02

Proof. Let us only prove (9.8), since the proof of (9.9) is similar. First, consider the case γ+ = +∞. Then, by assumption, there exists M > 0 such that Z zn+ < M, ∀n ∈ N. Ω o n 2M + For n ∈ N let Kn = x ∈ Ω : zn (x) < |Ω| . Then Z Z Z 2M 2M + + 0≤ zn = zn − zn+ ≤ M − (|Ω| − |Kn |) = |Kn | − M. |Ω| |Ω| Kn Ω Ω\Kn Therefore, |Kn | ≥ and

|Ω| , 2 µ

ku+ n kL2 (Kn ) teo:6

≤ |Kn |

1/2

sup γ

−1

2M |Ω|

¶ .

Then, by Proposition 95, for all n ∈ N, Ãµ Z µ ¶1/2 ¶! 2M + ˜ + |Ω|1/2 sup γ −1 − Au+ . ku+ n un n kL2 (Ω) ≤ C(J, Ω) |Ω| Ω

120


Now, let us consider the case γ+ < +∞. Let δ = γ+ |Ω| − M.

111n

By assumption, for every n ∈ N, we have, Z (9.10) zn < γ+ |Ω| − δ. Ω

n

o 111n δ For n ∈ N, let Kn = x ∈ Ω : zn (x) < γ+ − 2|Ω| . Then, by (9.10), ¶ µ Z Z Z δ δ . zn = zn − zn < − + |Kn | γ+ − 2 2|Ω| Kn Ω Ω\Kn Moreover,

Z

Z

Z

zn = − Kn

|zn | + Kn ∩{x∈Ω:zn 0, h −

δ 2|Ω|

+ γ+ > 0 and |Kn | ≥

³ 4 h−

δ δ 2|Ω|

+ γ+

Consequently,

µ ku+ n kL2 (Kn ) teo:6

≤ |Kn |

1/2

sup γ

−1

´.

δ γ+ − 2|Ω|

¶ .

Then, by Proposition 95, ku+ n kL2 (Ω)

Ãµ Z ¶1/2 µ ¶! δ + ˜ Ω, γ, h) ≤ C(J, − Au+ + |Ω|1/2 sup γ −1 γ+ − . n un 2|Ω| Ω her01

This ends the proof of (9.8).

¤

Finally, we have the following monotonicity result. Its proof is straightforward. lemma:m.II

Lemma 97. Let T : R → R a nondecreasing function. For every u ∈ L2 (Ω) such that T (u) ∈ L2 (Ω), it holds Z Z Z − Au(x)T (u(x)) dx = − J(x − y)(u(y) − u(x)) dy T (u(x)) dx Ω ZΩ ZΩ 1 = B(x, y)(u(y) − u(x))2 dy dx, 2 Ω Ω


121

where B(x, y) is the non negative symmetric function given by   J(x − y) T (u(y)) − T (u(x)) if u(y) 6= u(x), u(y) − u(x) B(x, y) =  0 if u(y) = u(x). In particular we have Z Z Z 1 J(x − y)(u(y) − u(x))2 dy dx. − Au(x) u(x) dx = 2 Ω Ω Ω 0.24. Mild solutions and contraction principle. In this section we obtain a mild solution to our problem studying the associated integral operator. Given a maximal monotone graph γ in R2 such that 0 ∈ γ(0), γ− < γ+ , we consider the problem, (Sφγ ) def:1

γ(u) − Au 3 φ

in Ω.

Definition 98. Let φ ∈ L1 (Ω). A pair of functions (u, z) ∈ L2 (Ω)×L1 (Ω) is a solution of problem (Sφγ ) if z(x) ∈ γ(u(x)) a.e. x ∈ Ω and z(x) − Au(x) = φ(x) a.e. x ∈ Ω, that is, Z z(x) −

J(x − y)(u(y) − u(x)) dy = φ(x)

a.e. x ∈ Ω.

Ω

With respect to uniqueness of problem (Sφγ ), we have the following maximum principle. th:Uniq.II

Theorem 99. (i) Let φ1 ∈ L1 (Ω) and (u1 , z1 ) a subsolution of (Sφγ1 ), that is, z1 (x) ∈ γ(u1 (x)) a.e. x ∈ Ω and z1 (x) − Au1 (x) ≤ φ1 (x) a.e. x ∈ Ω, and let φ2 ∈ L1 (Ω) and (u2 , z2 ) a supersolution of (Sφγ2 ), that is, z2 (x) ∈ γ(u2 (x)) a.e. x ∈ Ω and z2 (x) − Au2 (x) ≥ φ2 (x) a.e. x ∈ Ω. Then Z

Z +

(φ1 − φ2 )+ .

(z1 − z2 ) ≤ Ω

Ω

Moreover, if φ1 ≤ φ2 , φ1 6= φ2 , then u1 (x) ≤ u2 (x) a.e. x ∈ Ω. (ii) Let φ ∈ L1 (Ω), and (u1 , z1 ), (u2 , z2 ) two solutions of (Sφγ ). Then, z1 = z2 a.e. and there exists a constant c such that u1 = u2 + c, a.e. Proof. To prove (i), let (u1 , z1 ) a subsolution of (Sφγ1 ) and (u2 , z2 ) a supersolution of (Sφγ2 ). Then −(Au1 (x) − Au2 (x)) + z1 (x) − z2 (x) ≤ φ1 (x) − φ2 (x).

122

eq:100


Multiplying the above inequality by k1 Tk+ (u1 − u2 + k sign+ 0 (z1 − z2 )) and integrating we get, Z 1 (z1 − z2 ) Tk+ (u1 − u2 + k sign+ 0 (z1 − z2 )) k Ω Z 1 − (Au1 (x) − Au2 (x)) Tk+ (u1 (x) − u2 (x) + k sign+ (9.11) 0 (z1 (x) − z2 (x))) dx k Ω Z Z 1 ≤ (φ1 − φ2 ) Tk+ (u1 − u2 + k sign+ (φ1 (x) − φ2 (x))+ dx. 0 (z1 − z2 )) ≤ k Ω Ω Let us write u = u − u2 and z = sign+ 0 (z1 − z2 ), then, by the monotonicity proved in lemma:m.II 1 Lemma 97, Z 1 lim (Au1 (x) − Au2 (x)) Tk+ (u1 (x) − u2 (x) + k sign+ 0 (z1 (x) − z2 (x))) dx k→0 Ω k Z 1 = lim Au(x) Tk+ (u(x) + kz(x)) dx k→0 Ω k Z 1 = lim A(u + kz)(x) Tk+ (u(x) + kz(x)) dx ≤ 0. k→0 Ω k eq:100

Therefore, taking limit as k goes to 0 in (9.11), we obtain Z Z + (z1 − z2 ) ≤ (φ1 − φ2 )+ . Ω

Ω

Let us now suppose that φ1 ≤ φ2 , φ1 6= φ2 . By the previous calculations we know that z1 ≤ z2 . Since Z Z Z Z z1 ≤ φ1 < φ2 ≤ z2 , eq:100

Ω

Ω

Ω

Ω

z1 6= z2 . Going back to (9.11), if u = u1 − u2 , we get Z − Au(x) Tk+ (u(x)) dx = 0, Ω

and therefore,

Z −

Au(x) u+ (x) dx = 0.

lemma:m.IIΩ

Consequently, by Lemma 97, there exists a null set C ⊂ Ω × Ω such that blg:02

(9.12)

J(x − y)(u+ (y) − u+ (x))(u(y) − u(x)) = 0

for all (x, y) ∈ Ω × Ω \ C.

Let B a null subset of Ω such that if x ∈ / B, the section Cx = {y ∈ Ω : (x, y) ∈ C} is / B, if u(x) > 0 then, since there exists r0 > 0 such that J(z) > 0 for every null. Let x ∈ blg:02 z such that |z| ≤ r0 , by a compactness argument and having in mind (9.12), it is easy to see that u(y) = u(x) > 0 for all y ∈ / Cx . Therefore u1 (y) > u2 (y) for all y ∈ / Cx in Ω and consequently z1 (y) ≥ z2 (y) a.e. in Ω which contradicts that z1 ≤ z2 , z1 6= z2 .


123

Let us now prove (ii). As (ui , zi ) are solutions of (Sφγ ) we have that −(Au1 (x) − Au2 (x)) + z1 (x) − z2 (x) = 0. Now, by (i), z1 = z2 , a.e. Consequently, 0 = −(Au1 (x) − Au2 (x)) = −A(u1 − u2 )(x). Therefore, multiplying the above equation by u = u1 − u2 and integrating we obtain Z Z 1 J(x − y)(u(y) − u(x))2 dy dx = 0. 2 Ω Ω Imejor.c.Poincare.II From here, by (9.2), u is constant a.e. in Ω.

¤

In particular we have the following result. MP

Corollary 100. Let k > 0 and u ∈ L2 (Ω) such that ku − Au ≥ 0

a.e. in Ω,

then u ≥ 0 a.e. in Ω. of (S0γ ), where γ(r) = kr, and (0, 0) is a Proof. Since (u, ku) is a supersolution th:Uniq.II subsolution of (S0γ ), by Theorem 99, the result follows. ¤ To study the existence of solutions of problem (Sφγ ) we start with the following two lemmas. lem:1

Lemma 101. Assume γ : R → R is a nondecreasing Lipschitz continuous function with γ(0) = 0 and γ− < γ+ . Let φ ∈ C(Ω) such that γ− < φ < γ+ . Then, there exists a solution (u, γ(u)) of problem (Sφγ ). Moreover, γ(u) ¿ φ. Proof. Since γ− < φ < γ+ and φ ∈ C(Ω), we can find c1 ≤ c2 such that

MPE1

(9.13)

γ− < γ(c1 ) ≤ φ(x) ≤ γ(c2 ) < γ+

∀ x ∈ Ω.

Since γ is a nondecreasing Lipschitz continuous function there exists k > 0 for which the function s 7→ ks − γ(s) is nondecreasing. Let us see by induction that we can find a sequence {ui } ⊂ L2 (Ω) such that u0 = c 1 , MPE2

(9.14)

ui ≤ ui+1 ≤ c2 ,

kui+1 − Aui+1 = φ − γ(ui ) + kui ,

∀ i ∈ N.

Since k > 0, as a consequence of being A self-adjoint, it is easy to see that k does not belong to the spectrum of A, then there exists u1 ∈ L2 (Ω) such that MPE1

ku1 − Au1 = φ − γ(c1 ) + kc1 .

Then, by (9.13), we have ku1 − Au1 = φ − γ(c1 ) + kc1 ≥ kc1 = kc1 − Ac1 .

124


MP

Hence, from Corollary 100 we get that u0 = c1 ≤ u1 . Analogously, there exists u2 such that ku2 − Au2 = φ − γ(u1 ) + ku1 . Now, since c1 ≤ u1 , we get ku2 − Au2 ≥ φ − γ(c1 ) + kc1 = ku1 − Au1 . MP

Again by Corollary 100, we get u1 ≤ u2 , and by induction we obtain that ui ≤ u i+1 . On MPE1 the other hand, since the function s 7→ ks − γ(s) is nondecreasing, c1 ≤ c2 and (9.13), we have kc2 − Ac2 ≥ φ − γ(c2 ) + kc2 ≥ φ − γ(c1 ) + kc1 = ku1 − Au1 . MP

Applying again Corollary 100, weMPE2 get c2 ≥ u1 , and by an inductive argument we deduce that ui ≤ c2 for all i ∈ N. Hence (9.14) holds. Consequently, there exists u ∈ L∞ (Ω), such MPE2 that u(x) = limi→+∞ ui (x) a.e. in Ω. Taking limits in (9.14), we obtain that ku − Au = φ − γ(u) + ku, and (u, γ(u)) is a solution of problem (Sφγ ), that is, blg:03

(9.15)

γ(u) − Au = φ. blg:03

Finally, given p ∈ P0 , multiplying (9.15) by p(γ(u)), and integrating in Ω, we get Z Z Z γ(u(x))p(γ(u(x))) dx − Au(x)p(γ(u(x))) dx = φ(x)p(γ(u(x))) dx. Ω

lemma:m.II

Ω

Ω

Now, by Lemma 97, the second term in the above equality is nonnegative, therefore Z Z γ(u(x))p(γ(u(x))) dx ≤ φ(x)p(γ(u(x))) dx. 00II

Ω

By Proposition 92, we conclude that γ(u) ¿ φ. th:661

Ω

¤

Lemma 102. Assume γ is a maximal monotone graph in R2 , ]−∞, 0] ⊂ D(γ), 0 ∈ γ(0), γ− < γ+ . Let γ˜ (s) = γ(s) if s < 0, γ˜ (s) = 0 if s ≥ 0. Assume γ˜ is Lipschitz continuous in ] − ∞, 0]. Let φ ∈ C(Ω) such that γ− < φ < γ+ . Then, there exists a solution (u, z) of (Sφγ ). Moreover, z ¿ φ. Proof. If γ− < 0, let c1 such that γ(c1 ) = {m1 }, γ− < m1 < 0 and m1 ≤ φ. And if γ− = 0 let c1 = m1 = 0. Let γr , r ∈ N, be the Yosida approximation of γ and let the maximal monotone graph ( γ(s) if s < 0, γ r (s) = γr (s) if s ≥ 0. Observe that γ r is a nondecreasing Lipschitz continuous function with γ r (0) = 0 and, for r large enough, γ− = γ−r < φ < γ+r , γ r ≤ γ r+1 , and converges in the sense of maximal


125

monotone graphs to γ. From the previous lemma, for each γ r we obtain a solution (ur , zr ) r of (Sφγ ), that is, zr = γ r (ur ) a.e. and eq:102

(9.16)

zr − Aur = φ.

Moreover, zr ¿ φ, and consequently, zr ≥ m1 . Moreover, ur ≥ c1 . Let ( zr (x) if ur (x) ≤ 0, zˆr (x) = γr+1 (ur (x)) if ur (x) > 0. Then, since γr is nondecreasing, zˆr ≥ zr , and also, zˆr ∈ γ r+1 (ur ). Therefore, (ur , zˆr ) is a supersolution to (Sφγ

r+1

th:Uniq.II

). Using Theorem 99, we obtain that

zˆr ≥ zr+1 . Now, if zˆr = zr then zr ≥ zr+1 ,

th:Uniq.II

and if zˆr 6= zr , by Theorem 99,

ur ≥ ur+1 . So, there exists a monotone non increasing subsequence of {ur }, denoted equal, with ur ≥ cˆ1 , or there exists a monotone non increasing subsequence of {zr }, denoted equal, with zr ≥ m1 . In the first case, we have that ur → u in L2 (Ω), and also, since zr ¿ φ, weakly in L1 (Ω).

zr → z And in the second case, we obtain eq:103

(9.17)

zr → z

in L1 (Ω).

zr → z

in L2 (Ω).

In fact, since zr ¿ φ, we get that eq:103103

(9.18)

eq:102

Now, in this second case, multiplying (9.16) by ur − us and integrating we get Z Z Z − Aur (ur − us ) = φ (ur − us ) − zr (ur − us ). Ω

Moreover,

Ω

Z

Ω

Z

−

Z

Aus (ur − us ) = Ω

φ (ur − us ) − Ω

zs (ur − us ). Ω

126


Hence, since

R

R = Ω zs , Z Z − A(ur − us )(ur − us ) = − (zr − zs )(ur − us ) Ω µ µ ¶¶ Z Z Ω Z 1 1 = − (zr − zs ) ur − ur − us − us |Ω| Ω |Ω| Ω Ω

z Ω r

blg:01.II

and, by Proposition 93, °µ ¶ µ ¶° Z Z ° ° 1 1 ° β 1 ° ur − ur − us − us ° ° 2 ≤ kzr − zs kL2 (Ω) . |Ω| Ω |Ω| Ω L (Ω) eq:103103

From (9.18) we get, n

R

o

1 ur − |Ω|

Z ur → w

in L2 (Ω).

Ω

1 Let us see that |Ω| u is bounded. If not, we can assume, passing to a subsequence if Ω r necessary, that it converges to −∞. Then, ur R→ −∞Ra.e. in Ω. Since zr ∈ γ r (ur ), γ r → γ eq:103 and (9.17), z =nγ− a.e. in o Ω. Consequently, Ω φ = Ω z = |Ω|γ− which contradicts that R 1 φ > γ− . Thus, |Ω| Ω ur is bounded and we have that there exists a subsequence of {ur }, denoted equal, such that ur → u in L2 (Ω). Therefore, in both cases, z ∈ γ(u) a.e. in Ω, z ¿ φ, and, taking limit in

zr − Aur 3 φ, we obtain z − Au 3 φ, which concludes the proof.

¤

With this lemma in mind we proceed to extend the result for general monotone graphs. th:Exist

Theorem 103. Assume γ is a maximal monotone graph in R2 , 0 ∈ γ(0) and γ− < γ+ . Let φ ∈ C(Ω) such that γ− < φ < γ+ . Then, there exists a solution (u, z) of (Sφγ ). Moreover, z ¿ φ. Proof. Let γr , r ∈ N, be the Yosida approximation of γ and let the maximal monotone graph ( γ(s) if s > 0, γ r (s) = γr (s) if s ≤ 0. th:661

Observe that γ r satisfies the hypothesis of Lemma 102, γ−r < φ < γ+r for r large enough, γ r ≥ γ r+1 and converges in the sense of maximal monotone graphs to γ. From the previous r lemma, for each γ r we obtain a solution (ur , zr ) of (Sφγ ), zr ¿ φ. Now, we can proceed similarly to the previous lemma passing to the limit to conclude. ¤


127

The natural space to study the problem PγJ (z0 ) from the point of view of Nonlinear Semigroup Theory is L1 (Ω). In this space we define the following operator, © ª γ B γ := (z, zˆ) ∈ L1 (Ω) × L1 (Ω) : ∃u ∈ L2 (Ω) such that (u, z) is a solution of (Sz+ˆ z) ,

E1Oper0

in other words, zˆ ∈ B γ (z) if and only if there exists u ∈ L2 (Ω) such that z(x) ∈ γ(u(x)) a.e. in Ω, and Z (9.19) − J(x − y)(u(y) − u(x)) dy = zˆ(x), a.e. x ∈ Ω. Ω γ

CProblm0.II

The operator B allows us to rewrite PγJ (z0 ) as the following abstract Cauchy problem in L1 (Ω), ( 0 z (t) + B γ (z(t)) 3 0 t ∈ (0, T ) (9.20) z(0) = z0 . th:Uniq.II th:Exist

A direct consequence of Theorems 99 and 103 is the following result. cor:01

Corollary 104. Assume γ is a maximal monotone graph in R2 , 0 ∈ γ(0). Then, the operator B γ is T -accretive in L1 (Ω) and satisfies © ª φ ∈ C(Ω) : γ− < φ < γ+ ⊂ Ran(I + B γ ). The following theorem is a consequence of the above result.

EUWSCT

eq:91

Theorem 105. Let T > 0 and zi0 ∈ L1 (Ω), i = 1, 2. Let zi be a solution in [0, T ] of PγJ (zi 0 ), i = 1, 2. Then Z Z + (z1 (t) − z2 (t)) ≤ (z10 − z20 )+ (9.21) Ω

Ω

for almost every t ∈]0, T [. Proof. Let (ui (t), zi (t)) be solutions of eq:91 PγJ (z0 i ), i = 1, 2. Then, since they are strong CProblm0.II solutions BCP of (9.20) and A is T -accretive, (9.21) follows from the Nonlinear Semigroup Theory ([20]). ¤ L1 (Ω)

In the next result we characterize D(B γ ) th:2

.

Theorem 106. Assume γ is a maximal monotone graph in R2 . Then, we have ª © L1 (Ω) D(B γ ) = z ∈ L1 (Ω) : γ− ≤ z ≤ γ+ . Proof. It is obvious that L1 (Ω)

D(B γ )

ª © ⊂ z ∈ L1 (Ω) : γ− ≤ z ≤ γ+ .

To obtain the another inclusion, it is enough to take φ ∈ C(Ω), satisfying γ− < φ < γ+ , L1 (Ω)

and to prove that φ ∈ D(B γ )

. Let a, b ∈ R such that γ− < a < φ < b < γ+ .

128

E1Cont


¡ ¢−1 th:Exist Now, by Theorem 103, for any n ∈ N, there exists vn := I + n1 B γ φ ∈ D(B γ ). Then, γ 2 (vn , n(φ − vn )) ∈ B , thus there exists un ∈ L (Ω) such that vn ∈ γ(un ) a.e. in Ω and Z 1 (9.22) vn (x) − J(x − y)(un (y) − un (x)) dy = φ(x) ∀ x ∈ Ω. n Ω Moreover, vn ¿ φ. Then,

E2Cont

(9.23) E1Cont

−∞ < inf γ −1 (a) ≤ un ≤ sup γ −1 (b) < +∞. E2Cont

Hence, from (9.22) and (9.23) it follows that vn → φ in L1 (Ω).

¤

As a consequence of the above results we have the following theorem concerning mild BCP solutions (see [20]). th:3.II

Theorem 107. Assume γ is a maximal monotone graph in R2 . Let T > 0 and let CProblm0.II 1 z0 ∈ L (Ω) satisfying γ− ≤ z0 ≤ γ+ . Then, there exists a unique mild solution of (9.20). Moreover z ¿ z0 . Proof. For n ∈ N, let ε = T /n, and consider a subdivision t0 = 0 < t1 < · · · < tn−1 < T = tn with ti − ti−1 = ε. Let z0ε ∈ C(Ω) with γ− < z0ε < γ+ and kz0ε − z0 kL1 (Ω) ≤ ε.

th:Exist

By Theorem 103, for n large enough, there exists a solution (uεi , ziε ) of ProbR0

ε γ(uεi ) − εAuεi 3 zi−1

(9.24) for i = 1, . . . , n, with

OOProbR0

ε ziε ¿ zi−1 .

(9.25)

That is, there exists a unique solution ziε ∈ L1 (Ω) of the time discretized scheme associated CProblm0.II with (9.20), ε ziε + εB γ ziε 3 εzi−1 , for i = 1, . . . , n. Therefore, if we define zε (t) by ( zε (0) = z0ε , zε (t) = ziε ,

for t ∈]ti−1 , ti ], i = 1, . . . , n, CProblm0

it is an ε-approximate solution of problem (10.3). Be BCP Cr By using now the Nonlinear Semigroup Theory (see [18], [20], [49]), on account of cor:01 th:2 CProblm0.II Corollary 104 and Theorem 106, problem (9.20) has a unique mild-solution z(t) ∈ C([0,OOProbR0 T] : 1 1 L (Ω)), obtained as z(t) = L (Ω)-limε→0 zε (t) uniformly for t ∈ [0, T ]. Finally, from (9.25) we get z ¿ z0 . ¤


129

Cr

Italia

By Crandall-Liggett’s Theorem, [49], the mild solution obtained above is given by the well-known exponential formula, µ ¶−n t γ −tB γ (9.26) e z0 = lim I + B z0 . n→∞ n γ

The nonlinear contraction semigroup e−tB generated by the operator −B γ will be denoted in the sequel by (S(t))t≥0 . In principle, it is not clear how these mild solutions have to be interpreted respect to In the next section we will see that they coincide with the solutions defined in the Introduction.

PγJ (z0 ).

:existencia

0.25. Existence of solutions. In this section we prove that the mild solution of CProblm0.II (9.20) is in fact a solution of the problem PγJ (z0 ). R 1 Theorem 108. Let z0 ∈ L1 (Ω) such that γ− ≤ z0 ≤ γ+ , γ− < |Ω| z < γ+ and Ω 0 R ∗ J j (z ) < +∞. Then, there exists a unique solution to Pγ (z0 ) in [0, T ] for every T > 0. Ω γ 0 Moreover, z ¿ z0 . Proof. We divide the proof in three steps. Step 1. First, let us suppose that

ecstep1

(9.27)

there exist c1 , c2 such that c1 ≤ c2 , m1 ∈ γ(c1 ), m2 ∈ γ(c2 ) and γ− < m1 ≤ z0 ≤ m2 < γ+ . CProblm0.II

th:3.II

Let z(t) be the mild solution of (9.20) given by Theorem 107. We shall show that z is a solution of problem PγJ (z0 ). For n ∈ N, let ε = T /n, and consider a subdivision t0 = 0 < t1 < · · · < tn−1 < T = tn with ti − ti−1 = ε. Then, it follows that 2cumpl002

E1Dis2

E2Oper

(9.28)

z(t) = L1 (Ω)- limε zε (t)

uniformly for t ∈ [0, T ],

where zε (t) is given, for ε small enough, by ( zε (t) = z0 for t ∈] − ∞, 0], (9.29) zε (t) = zin , for t ∈]ti−1 , ti ], i = 1, . . . , n, where (uni , zin ) ∈ L2 (Ω) × L1 (Ω) is the solution of n zin − zi−1 n (9.30) −Aui + = 0, i = 1, 2, . . . , n. ε Moreover, zin ¿ z0 . Hence γ− < m1 ≤ zin ≤ m2 < γ+ and consequently, inf γ −1 (m1 ) ≤ uni ≤ sup γ −1 (m2 ). Therefore, if we write uε (t) = uni , t ∈]ti−1 , ti ], i = 1, . . . , n, we can suppose that

e1CONVer

(9.31)

uε * u weakly in L2 (0, T ; L2 (Ω)) as ε → 0+ .

130


e1CONVer

Since zε ∈ γ(uε ) a.e. in QT , zε → z in L1 (QE2Oper T ), having in mind (9.31), we obtain that z ∈ γ(u) a.e. in QT . On the other hand, from (9.30), zε (t) − zε (t − ε) * zt ε

weakly in L2 (0, T ; L2 (Ω)) as ε → 0+ .

Step 2. Let now z0 ∈ L1 (Ω) such that γ− ≤ z0 ≤ γ+ , γ− |Ω| < +∞, and ecstep2

R

z Ω 0

< γ− |Ω|,

R

j ∗ (z ) Ω γ 0

Ω z0 n+1 and z0 n ≥ m1 (n) > γ− , m1 (n) ∈ γ(c1 (n)) for some c1 (n). CProblm0.II By Step 2, there exist a solution zn of problem PγJ (z0 n ), which is the mild solution of (9.20) with initial datum z0 n and satisfies zn ¿ z0 . It is obvious that ecua:her03

(9.38)

in C([0, T ] : L1 (Ω)),

lim zn = z

n→∞

CProblm0.II

eq:57333

being z the mild solution of (9.20) with initial datum z0 . Moreover z ¿ z0 . We shall see that z is the solution of PγJ (z0 ). The proof is similar to the above step and we only need to take care in the proof of the boundedness of {un } in L2 (QT ). To this end we need a eq:57 formula like (9.37) for u− n , that is, we need to prove that there exists C > 0 such that Ãµ Z ! ¶1/2 (9.39) k(un (t))− kL2 (Ω) ≤ C − A(un (t))− (un (t))− +1 , ∀ t ∈ [0, T ]. Ω

Let us consider first that γ− = −∞, and let

Z z − (t) + 1.

M = sup t∈[0,T ]

Ω

Then, there exists n0 ∈ N such that Z sup (zn )− (t) < M, t∈[0,T ]

ecua:her01

In the case γ− > −∞, there exists M ∈ R, h > 0 and n0 ∈ N such that, for all n ≥ n0 , Z (9.40) inf zn (t) > M > γ− |Ω| t∈[0,T ]

and ecua:her02

∀n ≥ n0 .

Ω

Ω

Z

(9.41)

sup t∈[0,T ]

ecua:her01

zn (t) < {x∈Ω:zn (t)(x)>h}

M − γ− |Ω| . 4

ecua:her02

ecua:her03

ecua:her03

Formula (9.40) is straightforward and (9.41) follows from (9.38). Indeed, by (9.38), there exists n0 ∈ N, δ > 0 and h > 0 such that, for all n ≥ n0 and for all t ∈ [0, T ], Z M − γ− |Ω| , ∀E ⊂ Ω, |E| < δ, |zn (t)| < 4 E and we can take h satisfying |{x ∈ Ω : zn (t)(x) > h}| < δ. lemma:01 eq:57333

Therefore, in both cases, by Lemma 96, (9.39) is proved. EUWSCT

Uniqueness of solutions follows from Theorem 105.

¤


133

0.26. Asymptotic behaviour. In this section we study the asymptotic behaviour of the solutions to PγJ (z0 ). Note that since the solution preserves the total mass it is natural to expect that solutions to our diffusion problem converge to the mean value of the initial condition as t → ∞. We shall see that this is the case, for instance, when γ is a continuous function, nevertheless this fails when γ has jumps. Let us introduce the ω−limit set for a given initial condition z0 , © ª ω(z0 ) = w ∈ L1 (Ω) : ∃ tn → ∞ with S(tn )z0 → w, strongly in L1 (Ω) and the weak ω−limit set © ª ωσ (z0 ) = w ∈ L1 (Ω) : ∃ tn → ∞ with S(tn )z0 * w, weakly in L1 (Ω) . Since S(t)z0 ¿ z0 , ωσ (z0 ) 6= ∅ always. Moreover since S(t) preserves the total mass, for all w ∈ ωσ (z0 ), Z Z w= z0 . Ω

Ω

We denote by F the set of fixed points of the semigroup (S(t)), that is, o n L1 (Ω) : S(t)w = w ∀ t ≥ 0 . F = w ∈ D(B γ ) It is easy to see that EQUI

eo.asymp.II

(9.42)

© ª F = w ∈ L1 (Ω) : ∃k ∈ D(γ) such that w ∈ γ(k) .

R 1 z < γ+ and Theorem 109. Let z0 ∈ L1 (Ω) such that γ− ≤ z0 ≤ γ+ , γ− < |Ω| Ω 0 R ∗ j (z ) < +∞. Then, ωσ (z0 ) ⊂ F . Moreover, if ω(z0 ) 6= ∅, then ω(z0 ) consists of a Ω γ 0 unique w ∈ F , and consequently, lim S(t)z0 = w

t→∞

zidane

Proof. Along this proof we denote by z(t) = S(t)z0 the solution to problem PγJ (z0 ) and u(t) the corresponding function that appears in Definition 1.1. Multiplying the equation in PγJ (z0 ) by u(t) and integrating, we deduce Z +∞ Z Z (9.43) − Au(t) u(t) dt ≤ jγ∗ (z0 ). 0

blg:01.II

zuazua

strongly in L1 (Ω).

Ω

Ω

Therefore, thanks to (93), we obtain that there exists a constant C such that ¯2 Z Z +∞ Z ¯ ¯ ¯ 1 ¯ dt ≤ C. ¯ u(t) − u(t) (9.44) ¯ ¯ |Ω| Ω 0 Ω Let w ∈ ωσ (z0 ), then there exists a sequence tn → +∞ such that S(tn )z0 * w. By zuazua (9.44), we have ¯2 Z +∞ Z ¯ Z ¯ ¯ 1 ¯ αn := u(t)¯¯ dt → 0. ¯u(t) − |Ω| tn

Ω

Ω

134


Take sn → 0 such that cero

αn = 0. sn By contradiction it is easy to see that there exists t¯n ∈ [tn , tn + ¯2 Z ¯ Z ¯ ¯ ¯u(t¯n ) − 1 u(t¯n )¯¯ ≤ sn , ¯ |Ω| Ω Ω (9.45)

and consequently, ==BLGamba

(9.46)

lim

n→∞

C ] sn

such that

¯2 Z ¯ Z ¯ ¯ ¯u(t¯n ) − 1 ¯ → 0. ¯ u( t ) n ¯ ¯ |Ω| Ω Ω

Let us prove that

Z 1 u(t¯n ) |Ω| Ω is bounded. In fact, assume there exists a subsequence (still denoted by t¯n ) such that Z 1 u(t¯n ) → +∞. |Ω| Ω ==BLGamba By (9.46) we get that u(t¯n ) → +∞ a.e. Since z(t¯n ) ∈ γ(u(t¯n )), then z(t¯n ) → γ+ a.e. Moreover, as z(t¯n ) ¿ z0 and γ + ≥ 0, we can deduce that limn→∞ z(t¯n ) = limn→∞ z(t¯n )+ weakly in L1 (Ω). Hence, applying Fatou’s Lemma, we get Z Z z0 = lim z(t¯n )+ ≥ γ+ |Ω|, n→∞ Ω Ω Z 1 u(t¯n ) is bounded from below. Therea contradiction. A similar argument shows that |Ω| Ω fore, passing to a subsequence if necessary, we may assume that Z 1 u(t¯n ) → k |Ω| Ω ==BLGamba

==cordoba

for some constant k. Using again (9.46), (9.47) u(t¯n ) → k, strongly in L2 (Ω) and a.e. Since z(t¯n ) ¿ z0 , we can==cordoba assume, taking a subsequence if necessary, that z(t¯n ) * wˆ weakly 1 it follows that wˆ ∈ γ(k), and consequently w ˆ ∈ F . Let us in L (Ω). Then, from (9.47)cero show now that w = w. ˆ By (9.45), we have °Z ° °Z ° ° t¯n ° ° t¯n ° ° ° ° ° =° Au(s) ds ° kz(t¯n ) − z(tn )k = ° zt (s) ds ° ° tn ° 1 ° tn ° 1 L (Ω) L (Ω) ! ÃZ 1/2 ¯ ¯2 µ ¶1/2 Z +∞ Z ¯ ¯ α 1 n 1/2 ¯u(s) − ≤M C → 0, u(s)¯¯ ds ≤ M (t¯n − tn ) ¯ |Ω| Ω sn tn Ω


135

where M is a constant depending of |Ω|. Therefore, taking limit, we get z(t¯n ) − z(tn ) → 0, strongly in L1 (Ω), since it converges weakly to w − w, it follows that w = w, which is a fixed point. Finally, if ω(z0 ) 6= ∅, since ω(z0 ) ⊂ ωσ (z0 ) ⊂ F and (S(t)) is a contraction semigroup, we have that ω(z0 ) = {w} ⊂ F and lim S(t)z0 = w

t→∞

strongly in L1 (Ω). ¤

Remark 110. Note that in order to proof that ω(z0 ) 6= ∅, a usual tool is to show that the resolvent of B γ is compact. In our case this fails in general as the following example shows. Let γ any maximal monotone graph with γ(0) = [0, 1], zn ∈ L∞ (Ω), 0 ≤ zn ≤ 1 such that {zn } is not relatively compact in L1 (Ω). It is easy to check that zn = (I + B γ )−1 (zn ). Hence (I +B γ )−1 is not a compact operator in L1 (Ω). On the other hand, since the nonlocal operator does not have regularizing effects, here we cannot prove regularity properties of the solutions that would help to find compactness of the orbits. Nevertheless, we shall see in the next result that when γ is a continuous function we are able to prove that ω(z0 ) 6= ∅. Let us see now some cases in which ω(z0 ) 6= ∅ and Z 1 lim S(t)z0 = z0 strongly in L1 (Ω). t→∞ |Ω| Ω Given a maximal monotone graph γ in R × R, we set γ(r+) := inf γ(]r, +∞[),

cor.asymp

γ(r−) := sup γ(] − ∞, r[)

for r ∈ R, where we use the conventions inf ∅ = +∞ and sup ∅ = −∞. It is easy to see that γ(r) = [γ(r−), γ(r+)] ∩ R for r ∈ R. Moreover, γ(r−) = γ(r+) except at a countable set of points, which we denote by J(γ). R 1 Corollary 111. Let z0 ∈ L1 (Ω) such that γ− ≤ z0 ≤ γ+ , γ− < |Ω| z < γ+ and Ω 0 R ∗ j (z ) < +∞. The following statements hold. Ω γ 0 R R 1 1 z ∈ 6 γ(J(γ)) or z ∈ {γ(k+), γ(k−)} for some k ∈ J(γ), then (1) If |Ω| 0 |Ω| Ω 0 Ω Z 1 lim S(t)z0 = z0 strongly in L1 (Ω). t→∞ |Ω| Ω (2) If γ is a continuous function then Z 1 z0 strongly in L1 (Ω). lim S(t)z0 = t→∞ |Ω| Ω R 1 (3) If |Ω| z ∈]γ(k−), γ(k+)[ for some k ∈ J(γ), then Ω 0 ½ ¾ Z Z 1 ωσ (z0 ) ⊂ w ∈ L (Ω) : w ∈ [γ(k−), γ(k+)] a.e., w= z0 . Ω

Ω

136


and consequently, for any w ∈ ωσ (z0 ), there exists a non null set in which w ∈ ]γ(k−), γ(k+)[. Proof. (1). Along this proof we denote by z(t) = S(t)z0 the solution to problem PγJ (z0 ) and u(t)Rthe corresponding function that appears in Definition 1.1. First, let us 1 assume that |Ω| z 6∈ γ(J(γ)) and z0 ∈ L∞ (Ω). Working as in the above theorem, we Ω 0 have that there exists a constat k such that cordoba

(9.48)

u(tn ) → k,

strongly in L2 (Ω) and a.e.

Since z(tn ) ¿ z0 , there exists a subsequence such that z(tn ) * w weakly Rin L1 (Ω). RNow, 1 1 from z(tn ) ∈ γ(u(tn )) we deduce that w ∈ γ(k) and consequently, since |Ω| z = |Ω| w, Ω 0 Ω k 6∈ J(γ). Then, there exists δ > 0 such that γ is univalued and continuous on ]k − δ, k + δ[. Hence, w = γ(k) Rand z(tn ) → γ(k) a.e. Therefore, Since z(tn ) is bounded in L∞ (Ω), 1 z(tn ) → γ(k) = |Ω| z strongly in L1 (Ω). Then, by the above theorem we get that Ω 0 Z 1 z0 , as t → ∞. z(t) → |Ω| Ω The general case z0 ∈ L1 (Ω) follows easily from the previous arguments using again that we deal with a contraction semigroup. R 1 z ∈ {γ(k+), γ(k−)} for some k ∈ J(γ). It is easy to see that Assume now that |Ω| Ω 0 R ∗ R 1 1 j (z ) < +∞, we can find z0,n ∈ L (Ω), with γ− ≤ z0,n ≤ γ+ , γ− < |Ω| z < γ and 0,n + Ω γ 0,n RΩ 1 1 such that z0,n → z0 strongly in L (Ω) and verifying |Ω| Ω z0,n 6∈ γ(J(γ)) for all n. Then, by the above step, we have Z 1 S(t)z0,n → z0,n , strongly in L1 (Ω), |Ω| Ω from where it follows, using again that (S(t)) is a contraction semigroup, that Z 1 S(t)z0 → z0 , strongly in L1 (Ω). |Ω| Ω Statement (2) is an obvious consequence of (1) since in this case J(γ) = ∅. teo.asymp.II

Finally, we prove (3). Given w ∈ ωσ (z0 ), by Theorem 109, there exists k0 ∈ D(γ), such that w ∈ γ(k0 ). Then, k0 = k. In fact, if we assume, for instance, that k0 < k, then Z Z 1 1 γ(k0 +) ≥ w= z0 > γ(k−) > γ(k0 +), |Ω| Ω |Ω| Ω a contradiction. Hence, we have w ∈ γ(k), and Z Z 1 1 w= z0 ∈]γ(k−), γ(k+)[. |Ω| Ω |Ω| Ω Thus, w ∈ [γ(k−), γ(k+)] a.e. and, moreover, there exists a non null set in which w ∈ ]γ(k−), γ(k+)[. ¤


137

Remark 112. An alternative proof of the fact that ω(z0 ) ⊂ F is the following. Let Ψ : L1 (Ω) →] − ∞, +∞] the functional defined by

 Z  jγ∗ (z) Ψ(z) := Ω  +∞

if jγ∗ (z) ∈ L1 (Ω), if jγ∗ (z) 6∈ L1 (Ω).

Br2

Since jγ∗ is continuous and convex, Ψ is lower semi-continuous ([24], pag.160). Moreover, since S(t)z0 ¿ z0 for all t ≥ 0, we have Ψ(S(t)z0 ) ≤ Ψ(z0 ) for all t ≥ 0. Therefore, Ψ is a lower semi-continuous Liapunov functional for (S(t)). Then, by the Invariance Daf Principle of Dafermos ([51]), Ψ is constant on ω(z0 ). Consequently, given w0 ∈ ω(z0 ), if w(t) = S(t)w0 , we have Ψ(w(t)) is constant for eq:39 all t ≥ 0. Let u(t) such that w(t) ∈ γ(u(t)) and wt = A(u(t)). Working as in the proof of (9.35), we get Z Z Z d d d ∗ 0 = Ψ(w(t)) = j (w(t)) = jγ −1 (w(t)) = Au(t) u(t). dt dt Ω γ dt Ω Ω blg:01.II Then, by Proposition 93, we obtain that Z 1 u(t) = u(t). |Ω| Ω Hence, w(t) ∈ F for all t > 0, and consequently, w0 ∈ F .

CHAPTER 10

A non-local p−Laplacian with Neumann boundary conditions

p.lapla.Neu

Our main goal in this chapter is to study the following nonlocal nonlinear diffusion problem, which we call the nonlocal p-Laplacian problem (with homogeneous Neumann boundary conditions), Z   u (x, t) = J(x − y)|u(y, t) − u(x, t)|p−2 (u(y, t) − u(x, t)) dy, t PpJ (u0 ) Ω  u(x, 0) = u0 (x). Here J : RN R→ R is a nonnegative continuous radial function with compact support, J(0) > 0 and RN J(x)dx = 1 (this last condition is not necessary to prove the results of this chapter, it is imposed to simplify the exposition), 1 ≤ p < +∞ and Ω ⊂ RN is a bounded domain. When dealing with local evolution equations, two models of nonlinear diffusion has been extensively studied in the literature, the porous medium equation, ut = ∆um , and the p−Laplacian evolution, ut = div(|∇u|p−2 ∇u). In the first case (for the porous medium AMRT CER equation) a nonlocal analogous equation was studied in [7] (see also [40]). Our main objective in this paper is to study the nonlocal equation PpJ , that is, the nonlocal analogous to the p−Laplacian evolution. First, let us state the precise definition of a solution. Solutions to PpJ (u0 ) will be understood in the following sense. defdweakII

Definition 113. Let 1 < p < +∞. A solution of PpJ (u0 ) in [0, T ] is a function u ∈ C([0, T ]; L1 (Ω)) ∩ W 1,1 (]0, T [; L1 (Ω)) which satisfies u(0, x) = u0 (x) a.e. x ∈ Ω and Z ut (x, t) = J(x − y)|u(y, t) − u(x, t)|p−2 (u(y, t) − u(x, t)) dy a.e in ]0, T [×Ω. Ω

Let us note that, with this definition of solution, the evolution problem PpJ (u0 ) is the gradient flow associated to the functional Z Z 1 Jp (u) = J(x − y)|u(y) − u(x)|p dy dx, 2p Ω Ω which is the nonlocal analogous to the energy functional associated to the p−Laplacian Z 1 Fp (u) = |∇u(y)|p dy. p Ω 139

140

10. A NON-LOCAL p−LAPLACIAN WITH NEUMANN BOUNDARY CONDITIONS

Our first result shows existence and uniqueness of a global solution for this problem. Moreover, a contraction principle holds.

teo.1.intro

Theorem 114. Assume p > 1 and let u0 ∈ Lp (Ω). Then, there exists a unique solution defdweakII to PpJ (u0 ) in the sense of Definition 113. Moreover, if ui0 ∈ L1 (Ω), i = 1, 2, and ui is a solution in [0, T ] of PpJ (ui0 ). Then Z Z + (u1 (t) − u2 (t)) ≤ (u1 0 − u2 0 )+ for every t ∈]0, T [. Ω

Ω

p

If ui 0 ∈ L (Ω), i = 1, 2, then ku1 (t) − u2 (t)kLp (Ω) ≤ ku1 0 − u2 0 kLp (Ω)

for every t ∈]0, T [.

Let us finish the introduction by collecting some preliminaries and notations that will be used in the sequel. We denote by J0 and P0 the following sets of functions, J0 = {j : R → [0, +∞], convex and lower semi-continuos with j(0) = 0}, BC

/ supp(q)} . P0 = {q ∈ C ∞ (R) : 0 ≤ q 0 ≤ 1, supp(q 0 ) is compact, and 0 ∈

In [19] the following relation for u, v ∈ L1 (Ω) is defined, Z Z u ¿ v if and only if j(u) dx ≤ j(v) dx, Ω

Ω

and the following facts are proved. 00

Proposition 115. Let Ω be a bounded domain in RN . R R (i) For any u, v ∈ L1 (Ω), if Ω uq(u) ≤ Ω vq(u) for all q ∈ P0 , then u ¿ v. (ii) If u, v ∈ L1 (Ω) and u ¿ v, then kukr ≤ kvkr for any r ∈ [1, +∞]. (iii) If v ∈ L1 (Ω), then {u ∈ L1 (Ω) : u ¿ v} is a weakly compact subset of L1 (Ω). 0.27. Existence of solutions for the nonlocal problems. The case p > 1. We first study the problem PpJ (u0 ) from the point of view of Nonlinear Semigroup Theory. For this we introduce in L1 (Ω) the following operator associated with our problem.

def.oper

Definition 116. For 1 < p < +∞ we define in L1 (Ω) the operator BpJ by Z J Bp u(x) = − J(x − y)|u(y) − u(x)|p−2 (u(y) − u(x)) dy, x ∈ Ω. Ω


141

Remark 117. It is easy to see that, 1. BpJ is positively homogeneous of degree p − 1, 2. Lp−1 (Ω) ⊂ Dom(BpJ ), if p > 2. 3. for 1 < p ≤ 2, Dom(BpJ ) = L1 (Ω) and BpJ is closed in L1 (Ω) × L1 (Ω). We have the following monotonicity lemma, whose proof is straightforward. lemma:m

DESIGU1

Lemma 118. Let 1 < p < +∞, and let T : R → R a nondecreasing function. Then, (i) for every u, v ∈ Lp (Ω) such that T (u − v) ∈ Lp (Ω), it holds Z (BpJ u(x) − BpJ v(x))T (u(x) − v(x))dx = Ω Z Z 1 (10.1) J(x − y) (T (u(y) − v(y)) − T (u(x) − v(x))) × 2 Ω Ω ¡ ¢ × |u(y) − u(x)|p−2 (u(y) − u(x)) − |v(y) − v(x)|p−2 (v(y) − v(x)) dy dx. DESIGU1

(ii) Moreover, if T is bounded, (10.1) holds for u, v ∈ Dom(BpJ ). In the next result we prove that BpJ is completely accretive and verifies a range condition. In short, this means that for any φ ∈ Lp (Ω) there is a unique solution of the problem u + BpJ u = φ and the resolvent (I + BpJ )−1 is a contraction in Lq (Ω) for all 1 ≤ q ≤ +∞. th:Uniq ecuacion01

Theorem 119. For 1 < p < +∞, the operator BpJ is completely accretive and verifies the range condition Lp (Ω) ⊂ Ran(I + BpJ ).

(10.2)

lemma:m

Proof. Given ui ∈ Dom(BpJ ), i = 1, 2 and q ∈ P0 , by the monotonicity Lemma 118, we have Z (BpJ u1 (x) − BpJ u2 (x))q(u1 (x) − u2 (x)) dx ≥ 0, Ω

BC

from where it follows that BpJ is a completely accretive operator (see [19]). To show that BpJ satisfies the range condition we have to prove that for any φ ∈ Lp (Ω) there exists u ∈ Dom(BpJ ) such that u = (I + BpJ )−1 φ. Let us first take φ ∈ L∞ (Ω). Let 0 An,m : Lp (Ω) → Lp (Ω) the continuous monotone operator defined by 1 1 An,m (u) := Tc (u) + BpJ u + |u|p−2 u+ − |u|p−2 u− . n m p We have that An,m is coercive in L (Ω). In fact, Z An,m (u)u Ω lim = +∞. kukLp (Ω) →+∞ kukLp (Ω)

142


Br1

Then, by Corollary 30 in [26], there exists un,m ∈ Lp (Ω), such that 1 1 Tc (un,m ) + BpJ un,m + |un,m |p−2 u+ |un,m |p−2 u− n,m − n,m = φ. n m 1 p−2 − Using the monotonicity of BpJ un,m + n1 |un,m |p−2 u+ un,m , we obtain that n,m − m |un,m | Tc (un,m ) ¿ φ. Consequently, taking c > kφkL∞ (Ω) , un,m ¿ φ and 1 1 |un,m |p−2 u+ |un,m |p−2 u− n,m − n,m = φ. n m Moreover, since un,m is increasing in n and decreasing in m. As un,m ¿ φ, we can pass to the limit as n → ∞ (using the monotone convergence to handle the term BpJ un,m ) obtaining um is a solution to 1 um + BpJ um − |um |p−2 u− m = φ. m Using um is decreasing in m we can pass again to the limit and to obtain un,m + BpJ un,m +

u + BpJ u = φ. Let now φ ∈ Lp (Ω). Take φn ∈ L∞ (Ω), φn → φ in Lp (Ω). Then, by our previous step, there exists un = (I + BpJ )−1 φn , un ¿ φn . Since BpJ is completely accretive, un → u in 0 Lp (Ω), also BpJ un → BpJ u in Lp (Ω) and we conclude that u + BpJ u = φ. ¤ th:Uniq

If BpJ denotes the closure of BpJ in L1 (Ω), then by Theorem 119 we obtain BpJ is mcompletely accretive in L1 (Ω). BCP

BC

As a consequence of the above results we get the following theorems (see [20] and [19]), teo.1.intro from which Theorem 114 can be derived. th:3

CProblm0 thth:3

Theorem 120. Assume p > 1. Let T > 0 and let u0 ∈ L1 (Ω). Then, there exists a unique mild solution u of ( 0 u (t) + BpJ u(t) = 0, t ∈ (0, T ), (10.3) u(0) = u0 . Theorem 121. Assume p > 1. Let T > 0. CProblm0

(1) Let u0 ∈ Lp (Ω). Then, the unique mild solution u of (10.3) is a solution of PpJ (u0 ) defdweakII in the sense of Definition 113. If 1 < p ≤ 2, this is true for any u0 ∈ L1 (Ω). (2) Let ui0 ∈ L1 (Ω), i = 1, 2, and ui a solution in [0, T ] of PpJ (ui0 ), i = 1, 2. Then Z Z + (u1 (t) − u2 (t)) ≤ (u1 0 − u2 0 )+ for every t ∈]0, T [. Ω

Ω

Moreover, for q ∈ [1, +∞], if ui 0 ∈ Lq (Ω), i = 1, 2, then ku1 (t) − u2 (t)kLq (Ω) ≤ ku1 0 − u2 0 kLq (Ω)

for every t ∈]0, T [.

.convergto1

ef.weak.p=1


143

Proof. The result follows from the fact that u(t) is a solution of P J (u ) if and only CProblm0 p 0 if u(t) is a strong solution of the abstract Cauchy problem (10.3). Now, u(t) is a strong solution under the hypothesis of the theorem thanks to the completely accretivity of BpJ ecuacion01 and the range condition (10.2). Moreover, the result follows for 1 < p ≤ 2, since in this case Dom(BpJ ) = L1 (Ω) and BpJ is closed in L1 (Ω) × L1 (Ω). ¤ Remark 122. Observe that our results can be extended (with minor modifications) to obtain existence and uniqueness for Z   u (x, t) = J(x, y)|u(y, t) − u(x, t)|p−2 (u(y, t) − u(x, t)) dy, t Ω  u(x, 0) = u0 (x), with J symmetric, that is, J(x, y) = J(y, x), bounded and nonnegative. The case p = 1 This chapter deals with the existence and uniqueness of solutions for the nonlocal 1-Lapla-cian problem with homogeneous Neumann boundary conditions, Z  u(y, t) − u(x, t)  u (x, t) = J(x − y) dy. t J |u(y, t) − u(x, t)| P1 (u0 ) Ω  u(x, 0) = u0 (x). We have that the formal evolution problem Z u(y, t) − u(x, t) ut (x, t) = J(x − y) dy, |u(y, t) − u(x, t)| Ω is the gradient flow associated to the functional Z Z 1 J1 (u) = J(x − y)|u(y) − u(x)| dy dx, 2 Ω Ω which is the nonlocal analogous to the energy functional associated to the total variation Z F1 (u) = |∇u(y)| dy. Ω

For p = 1 we give the following definition of what we understand as a solution. Definition 123. A solution of P1J (z0 ) in [0, T ] is a function u ∈ C([0, T ]; L1 (Ω)) ∩ W 1,1 (]0, T [; L1 (Ω)) which satisfies u(0, x) = u0 (x) a.e. x ∈ Ω and Z ut (x, t) = J(x − y)g(x, y, t) dy Ω

a.e in ]0, T [×Ω,

144


for some g ∈ L∞ (0, T ; L∞ (Ω × Ω)) with kgk∞ ≤ 1 such that g(x, y, t) = −g(y, t, x) and J(x − y)g(x, y, t) ∈ J(x − y)sign(u(y, t) − u(x, t)). To get existence and uniqueness of these kind of solutions, the idea is to take the limit as p & 1 of solutions to PpJ with p > 1.

teo.2.intro

Theorem 124. Assume p = 1 and let u0 ∈ L1 (Ω). Then, there exists a unique solution def.weak.p=1 to P1J (u0 ) in the sense of Definition 123. Moreover, for i = 1, 2, let ui 0 ∈ L1 (Ω) and ui be a solution in [0, T ] of P1J (ui 0 ). Then Z Z + (u1 (t) − u2 (t)) ≤ (u1 0 − u2 0 )+ for almost every t ∈]0, T [. Ω

Ω

As in the case p > 1, to prove the existence and uniqueness of solutions of P1J (u0 ) we use the Nonlinear Semigroup Theory, so we start introducing the following operator in L1 (Ω). pdef.oper2

Definition 125. We define the operator B1J in L1 (Ω) × L1 (Ω) by uˆ ∈ B1J u if and only if u, uˆ ∈ L1 (Ω), there exists g ∈ L∞ (Ω×Ω), g(x, y) = −g(y, x) for almost all (x, y) ∈ Ω×Ω, kgk∞ ≤ 1, Z uˆ(x) = −

J(x − y)g(x, y) dy,

a.e. x ∈ Ω

Ω

and ecua:07

(10.4)

J(x − y)g(x, y) ∈ J(x − y) sign(u(y) − u(x))

Remark 126.

a.e. (x, y) ∈ Ω × Ω.

ecua:07

1. It is not difficult to see that (10.4) is equivalent to Z Z Z Z 1 J(x − y)|u(y) − u(x)| dy dx, − J(x − y)g(x, y) dy u(x) dx = 2 Ω Ω Ω Ω 2. L1 (Ω) = Dom(B1J ) and B1J is closed in L1 (Ω) × L1 (Ω). 3. B1J is positively homogeneous of degree zero, that is, if uˆ ∈ B1J u and λ > 0 then uˆ ∈ B1J (λu). pth:Uniq

Theorem 127. The operator B1J is completely accretive and satisfies the range condition L∞ (Ω) ⊂ Ran(I + B1J ). Proof. Let uî ∈ B1J ui , i = 1, 2. Then there exists gi ∈ L∞ (Ω × Ω), kgi k∞ ≤ 1, gi (x, y) = −gi (y, x), J(x − y)gi (x, y) ∈ J(x − y)sign(ui (y) − ui (x)) for almost all (x, y) ∈ Ω × Ω, such that Z uî (x) = −

J(x − y)gi (x, y) dy, Ω

a.e. x ∈ Ω,


145

for i = 1, 2. Then, given q ∈ P0 , we have Z (ˆ u1 (x) − uˆ2 (x))q(u1 (x) − u2 (x)) dx Ω Z Z 1 = J(x − y)(g1 (x, y) − g2 (x, y)) (q(u1 (y) − u2 (y)) − q(u1 (x) − u2 (x))) dxdy 2 ZΩ Z Ω 1 = J(x − y)(g1 (x, y) − g2 (x, y))× 2 {(x,y):u1 (y)6=u1 (x),u2 (y)=u2 (x)} × (q(u1 (y) − u2 (y)) − q(u1 (x) − u2 (x))) dxdy Z Z 1 + J(x − y)(g1 (x, y) − g2 (x, y))× 2 {(x,y):u1 (y)=u1 (x),u2 (y)6=u2 (x)} × (q(u1 (y) − u2 (y)) − q(u1 (x) − u2 (x))) dxdy Z Z 1 + J(x − y)(g1 (x, y) − g2 (x, y))× 2 {(x,y):u1 (y)6=u1 (x),u2 (y)6=u2 (x)} × (q(u1 (y) − u2 (y)) − q(u1 (x) − u2 (x))) dxdy, and the last three integrals are nonnegative. Hence Z (ˆ u1 (x) − uˆ2 (x))q(u1 (x) − u2 (x)) dx ≥ 0, Ω

from where it follows that B1J is a completely accretive operator. To show that B1J satisfies the range condition, let us see that for any φ ∈ L∞ (Ω), lim (I + BpJ )−1 φ = (I + B1J )−1 φ weakly in L1 (Ω).

p→1+

=pecu:01

¡ ¢−1 Let φ ∈ L∞ (Ω), and write, for 1 < p < +∞, up = I + BpJ φ. Then, Z up (x) − J (x − y) |up (y) − up (x)|p−2 (up (y) − up (x)) dy = φ(x) a.e. x ∈ Ω. (10.5) Ω

Thus, for every v ∈ L∞ (Ω), we can write Z Z Z Z p−2 up v − J (x − y) |up (y) − up (x)| (up (y) − up (x)) dy v(x) dx = φv. Ω

Ω Ω

Ω

00

Since up ¿ φ, by Proposition 115, we have that there exists a sequence pn → 1 such that upn * u weakly in L1 (Ω), Observe that kupn kL∞ (Ω) , kukL∞ (Ω) ≤ kφkL∞ (Ω) .

u ¿ φ.

146


Now, since Z Z − J (x − y) |upn (y) − upn (x)|pn −2 (upn (y) − upn (x)) dy v(x) dx ΩZ ΩZ 1 = J(x − y) |upn (y) − upn (x)|pn −2 (upn (y) − upn (x)) (v(y) − v(x)) dy dx, 2 Ω Ω =pecu:01

taking v = upn , by (10.5), we get that Z Z 1 J(x − y) |upn (y) − upn (x)|pn dy dx ≤ M1 , 2 Ω Ω

∀ n ∈ N.

Therefore, for any measurable subset E ⊂ Ω × Ω, we have ¯Z Z ¯ ¯ ¯ p −2 n ¯ ¯ J(x − y)|u (y) − u (x)| (u (y) − u (x)) p p p p n n n n ¯ ¯ E Z Z 1 ≤ J(x − y)|upn (y) − upn (x)|pn −1 ≤ M2 |E| pn . E

Hence, by the Dunford-Pettis Theorem we may assume that there exists g(x, y) such that J(x − y)|upn (y) − upn (x)|pn −2 (upn (y) − upn (x)) * J(x − y)g(x, y), weakly in L1 (Ω × Ω), g(x, y) = −g(y, x) for almost all (x, y) ∈ Ω × Ω, and kgk∞ ≤ 1. =pecu:01

pecu:0101

Therefore, passing to the limit in (10.5) for p = pn , we get Z Z Z Z uv − J(x − y)g(x, y) dy v(x) dx = φv (10.6) Ω

Ω

Ω

Ω

∞

for every v ∈ L (Ω), and consequently we get Z u(x) − J(x − y)g(x, y) dy = φ(x)

a.e. x ∈ Ω.

Ω

pecua:02

Then, to finish the proof we have to show that Z Z Z Z 1 (10.7) − J(x − y)g(x, y) dy u(x) dx = J(x − y)|u(y) − u(x)| dy dx. 2 Ω Ω Ω Ω pecu:0101

In fact, by (10.6) with v = u, Z Z 1 J(x − y) |upn (y) − upn (x)|pn dy dx 2 ZΩ Ω Z Z Z Z =

φupn − upn upn = φu − uu − φ(u − upn ) Ω Ω Ω Ω Z Z + 2u(u − upn ) − (u − upn )(u − upn ) Ω Z ZΩ Z Z ≤− J(x − y)g(x, y) dy u(x) dx − φ(u − upn ) + 2u(u − upn ), Ω

Ω

Ω

Ω

Ω


147

so, 1 lim sup n→+∞ 2

Z Z

Z Z J(x − y) |upn (y) − upn (x)| Ω

pn

dy dx ≤ −

Ω

J(x − y)g(x, y) dy u(x) dx. Ω

Ω

lemma:m

Now, by the monotonicity Lemma 118, Z Z − J(x − y)|ρ(y) − ρ(x)|pn −2 (ρ(y) − ρ(x)) dy (upn (x) − ρ(x)) dx ΩZ ΩZ ≤− J(x − y)|upn (y) − upn (x)|pn −2 (upn (y) − upn (x)) dy (upn (x) − ρ(x)) dx. Ω

Ω

Therefore, taking limits, Z Z − J(x − y) sign0 (ρ(y) − ρ(x)) dy (u(x) − ρ(x)) dx Ω Ω Z Z J(x − y)g(x, y) dy (u(x) − ρ(x)) dx. ≤− Ω

Ω

pecua:02

Taking now, ρ = u ± λu, λ > 0, and letting λ → 0, we get (10.7), and the proof is finished. ¤ BCP

As a consequence of the above results we have the following theorems (see [20]), from teo.2.intro which Theorem 124 follows. 1th:3

1CProblm0 1thth:3

Theorem 128. Let T > 0 and u0 ∈ L1 (Ω). Then, there exists a unique mild solution u of ( 0 t ∈ (0, T ), u (t) + B1J u(t) 3 0, (10.8) u(0) = u0 . Theorem 129. Let T > 0. 1CProblm0

(1) Let u0 ∈ L1 (Ω). Then, the def.weak.p=1 unique mild solution u of (10.8) is a solution of P1J (u0 ) in the sense of Definition 123. (2) Let ui0 ∈ L1 (Ω), i = 1, 2, and ui a solution in [0, T ] of P1J (ui0 ), i = 1, 2. Then Z Z + (u1 (t) − u2 (t)) ≤ (u10 − u20 )+ for almost every t ∈]0, T [. Ω

Ω

0.28. Convergence to the p-laplacian. Our next step is to rescale the kernel J appropriately and take the limit as the scaling parameter goes to zero. To be more precise, for every p ≥ 1, we consider the local

148


p−Laplace evolution equation with homogeneous Neumann boundary conditions  ut = ∆ p u in ]0, T [×Ω,    |∇u|p−2 ∇u · η = 0 on ]0, T [×∂Ω, Np (u0 )    u(x, 0) = u0 (x) in Ω, where η is the unit outward normal on ∂Ω, ∆p u = div(|∇u|p−2 ∇u) is the p-laplacian of u. We obtain that the solutions of this local problem, Np (u0 ), can be approximated by solutions of a sequence of nonlocal p-Laplacian problems of the form PpJ . Problem N (u ), that is, the Neumann problem for the Total Variation Flow, was ABCM11 0 ACMBook studied in [2] (see also [3]), motivated by problems in image processing. This PDE appears when one uses the steepest descent method to RuOsFa minimize the Total Variation, a method introduced by L. Rudin, S. Osher and E. Fatemi [77] in the context of image denoising and reconstruction. Then, solving N1 (u0 ) amounts to regularize or, in other words, to filter the initial datum u0 . This filtering process has less destructive effect on the edges than filtering with a Gaussian, i.e., than solving the heat equation with initial condition u0 . In this context the given image u0 is a function defined on a bounded, smooth or piecewise smooth open subset Ω of RN , typically, Ω will be a rectangle in R2 . KOJ

funct1

S. Kindermann, S. Osher and P. W. Jones in [70] have studied deblurring andBCM denoising of images by nonlocal functionals, motivated by the use of neighborhood filters [29]. Such Y YE filters have originally been proposed by Yaroslavsky, [85], [86], and further generalized by TM KOJ C. Tomasi and R. Manduchi, [83], as bilateral filter. The main aim of [70] is to relate the neighborhood filter to an energy minimization. Now in this case the Euler-Lagrange equations are not partial differential equations but include integrals. The functional considered KOJ in [70] takes the general form µ ¶ Z |u(x) − u(y)|2 (10.9) Jg (u) = g w(|x − y|) dxdy, h2 Ω×Ω with w ∈ L∞ (Ω), g ∈ C 1 (R+ ) and h > 0 is a parameter. The Fréchet derivative of Jg as a functional from L2 (Ω) into R is given by µ ¶ Z |u(x) − u(y)|2 4 0 0 Jg (u)(x) = 2 g (u(x) − u(y))w(|x − y|) dy. h Ω h2 funct1

p

1 Note that the nonlocal functional Jp is of the form (10.9) with g(t) = 2p |t| 2 , w = J and h = 1. Then, problem PpJ (u0 ) appears when one uses the steepest descent method to minimize this particular nonlocal functional.

For given p ≥ 1 and J we consider the rescaled kernels CJ,p ³ x ´ Jp,ε (x) := p+N J , ε ε

onvergencia


149

where

Z 1 := J(z)|zN |p dz 2 RN is a normalizing constant in order to obtain the p-Laplacian in the limit instead a multiple of it. Associated with these rescaled kernels we have solutions uε to the equation in PpJ with J J replaced by Jp,ε and the same initial condition u0 (we shall call this problem Pp p,ε ). Our next result states that these functions uε converge strongly in Lp (Ω) to the solution to the local p−Laplacian Np (u0 ). −1 CJ,p

Theorem 130. Let Ω be a smooth bounded domain in RN and p ≥ 1. Assume J(x) ≥ J J(y) if |x| ≤ |y|. Let T > 0, u0 ∈ Lp (Ω) and uε the unique solution of Pp p,ε (u0 ). Then, if u is the unique solution of Np (u0 ), lim sup kuε (t, .) − u(t, .)kLp (Ω) = 0.

ε→0 t∈[0,T ]

Note that the above result states that PpJ is a nonlocal analogous to the p−Laplacian. Recall that for the linear case, p = 2, under additional regularity hypothesis on the involved data, the convergence of the solutions of rescaled nonlocal problems of the form sec.linear.neu CERW2 J P2 to the solution of the heat equation is proved in Chapter 5, see also [43]. Convergence to the p-laplacian for p > 1 Our main goal in this chapter is to show that the Neumann problem for the p-Laplacian equation Np (u0 ) can be approximated by suitable nonlocal Neumann problems PpJ (u0 ). Let us start recalling some results about the p-Laplacian equation  in ]0, T [×Ω,   ut = ∆ p u  |∇u|p−2 ∇u · η = 0 on ]0, T [×∂Ω, Np (u0 )    u(x, 0) = u0 (x) in Ω, AMSTAMSTTrans AIMTp

obtained in [5], [6] and [4]. We have the two following concepts of solutions. A weak solution of Np (u0 ) in [0, T ] is a function u ∈ C([0, T ] : L1 (Ω)) ∩ Lp (0, T ; W 1,p (Ω)) ∩ W 1,1 (0, T ; L1 (Ω)) with u(0) = u0 , satisfying Z Z 0 u (t)ξ + |∇u(t)|p−2 ∇u(t) · ∇ξ = 0 Ω

Ω

for any ξ ∈ W 1,p (Ω) ∩ L∞ (Ω).

for almost all t ∈]0, T [

150


An entropy solution of Np (u0 ) in the time interval [0, T ] is a function u ∈ C([0, T ] : L1 (Ω)) ∩ W 1,1 (0, T ; L1 (Ω)), with Tk (u) ∈ Lp (0, T ; W 1,p (Ω)) for all k > 0 such that u(0) = u0 and Z Z 0 u (t)Tk (u(t) − ξ) + |∇u(t)|p−2 ∇u(t) · ∇Tk (u(t) − ξ) = 0 for almost all t ∈]0, T [ Ω

for any ξ ∈ W

1,p

Ω ∞

(Ω) ∩ L (Ω).

Here the truncature functions Tk are defined by Tk (r) = k ∧ (r ∨ (−k)), k ≥ 0, r ∈ R. parEUWS

real.madrid

AMSTTrans AIMTp

Theorem 131 ([6], [4]). Let T > 0. For any u0 ∈ L1 (Ω) there exists a unique entropy 0 solution u(t) of Np (u0 ). Moreover, if u0 ∈ Lp (Ω) ∩ L2 (Ω) the entropy solution u(t) is a weak solution. Let us perform a formal calculation just to convince the reader that the convergence teo.3.intro.convergencia result, Theorem 130, is correct. Let N = 1. Let u(x) be a smooth function and consider ¶ Z µ x−y 1 Aε (u) = p+1 J |u(y) − u(x)|p−2 (u(y) − u(x)) dy. ε ε R Changing variables, y = x − εz, we get Z 1 (10.10) Aε (u) = p J(z)|u(x − εz) − u(x)|p−2 (u(x − εz) − u(x)) dz. ε R Now, we expand in powers of ε to obtain ¯ ¯ u00 (x) p−2 p−2 ¯ 0 2 2 ¯ |u(x − εz) − u(x)| = ε ¯u (x)z + 2 εz + O(ε )¯ = εp−2 |u0 (x)|p−2 |z|p−2 + εp−1 (p − 2)|u0 (x)z|p−4 u0 (x)z and u(x − εz) − u(x) = εu0 (x)z + real.madrid

u00 (x) 2 z + O(εp ), 2

u00 (x) 2 2 ε z + O(ε3 ). 2

Hence, (10.10) becomes Z 1 J(z)|z|p−2 z dz|u0 (x)|p−2 u0 (x) Aε (u) = ε RZ ¡ ¢ 1 + J(z)|z|p dz (p − 2)|u0 (x)|p−2 u00 (x) + |u0 (x)|p−2 u00 (x) + O(ε). 2 R Using that J is radially symmetric, the first integral vanishes and therefore, lim Aε (u) = C(|u0 (x)|p−2 u0 (x))0 ,

ε→0

where

1 C= 2

Z J(z)|z|p dz. R


151

To do this formal calculation rigorous we need to obtain the following result which is a BBM variant of [23, Theorem 4]. prop:01

new:05

Proposition 132. Let 1 ≤ q < +∞. Let ρ : RN → R be a nonnegative continuous radial function with compact support, non-identically zero, and ρn (x) := nN ρ(nx). Let {fn } be a sequence of functions in Lq (Ω) such that Z Z 1 (10.11) |fn (y) − fn (x)|q ρn (y − x) dx dy ≤ M q . n Ω Ω 1. If {fn } is weakly convergent in Lq (Ω) to f then (i) if q > 1, f ∈ W 1,q (Ω), and moreover ¢ µ ¶ ¡ fn x + n1 z − fn (x) 1 1/q χ (ρ(z)) z * (ρ(z))1/q z · ∇f Ω x+ n 1/n weakly in Lq (Ω) × Lq (RN ). (ii) If q = 1, f ∈ BV (Ω), and moreover ¢ ¶ ¡ µ fn x + n1 z − fn (x) 1 ρ(z)χΩ x + z * ρ(z)z · Df n 1/n weakly as measures. 2. Assume Ω is a smooth bounded domain in RN and ρ(x) ≥ ρ(y) if |x| ≤ |y|. Then {fn } is relatively compact in Lq (Ω), and consequently, there exists a subsequence {fnk } such that (i) if q > 1, fnk → f in Lq (Ω) with f ∈ W 1,q (Ω), (ii) if q = 1, fnk → f in L1 (Ω) with f ∈ BV (Ω). new:05

xcecu:02

e1

Proof. We suppose fn → f weakly in Lq (Ω) and write (10.11) as ¯ ¯ Z Z ¯ fn (y) − fn (x) ¯q N ¯ dx dy n ρ(n(x − y)) ¯¯ ¯ 1/n Ω Ω ¯q ¢ µ ¶ ¯¯ ¡ Z Z (10.12) ¯ 1 f x + z − f (x) 1 n ¯ n ¯ n = ρ(z)χΩ x + z ¯ ¯ dx dz ≤ M. ¯ ¯ n 1/n RN Ω On the other hand, if ϕ ∈ D(Ω) and ψ ∈ D(RN ), taking n large enough, ¶ µ Z Z fn (x + 1/nz) − fn (x) 1 1/q χΩ x + z (ρ(z)) ϕ(x)dxψ(z)dz n 1/n RN Ω ¡ ¢ Z Z fn x + n1 z − fn (x) 1/q (10.13) = (ρ(z)) ϕ(x)dxψ(z)dz 1/n RN Ω ¡ ¢ Z Z ϕ(x) − ϕ x − n1 z 1/q =− (ρ(z)) fn (x) dxψ(z)dz. 1/n RN Ω

152


xcecu:02

Let us start with the case 1.(i). By (10.12), up to a subsequence, ¢ µ ¶ ¡ fn x + n1 z − fn (x) 1 1/q χ (ρ(z)) z * (ρ(z))1/q g(x, z) Ω x+ n 1/n e1

weakly in Lq (Ω) × Lq (RN ). Therefore, passing to the limit in (10.13), we get Z Z Z Z 1/q 1/q (ρ(z)) f (x) z · ∇ϕ(x)dxψ(z)dz. (ρ(z)) g(x, z)ϕ(x) dxψ(z) dz = − RN

RN

Ω

Consequently, Z

Ω

Z g(x, z)ϕ(x)dx = −

f (x) z · ∇ϕ(x)dx ∀z ∈ int(supp(J)).

Ω

Ω

And from here, for s small, Z Z ∂ g(x, sei )ϕ(x)dx = − f (x) s ϕ(x)dx, ∂xi Ω Ω which implies f ∈ W 1,q (Ω) and (ρ(z))1/q g(x, z) = (ρ(z))1/q z · ∇f (x). xcecu:02

Let us now prove 1.(ii). By (10.12), there exists a bounded Radon measure µ ∈ M(Ω × RN ) such that, up to a subsequence, ¢ µ ¶ ¡ fn x + n1 z − fn (x) 1 ρ(z)χΩ x + z * µ(x, z) n 1/n e1

e2

weakly in M(Ω × RN ). Hence, passing to the limit in (10.13), we get Z Z (10.14) ϕ(x)ψ(z)dµ(x, z) = − ρ(z)ψ(z) z · ∇ϕ(x)f (x) dxdz. Ω×RN

Ω×RN

Ambrosio

Now, applying the disintegration theorem (Theorem 2.28 in [1]) to the measure µ, we get that if π : Ω × RN → RN is the projection on the first factor and ν = π# |µ|, then there exists a Radon measures µx in RN such that x 7→ µx is ν-measurable, |µx |(RN ) ≤ 1

ν − a.e.

in Ω

and, for any h ∈ L1 (Ω × RN , |µ|), h(x, ·) ∈ L1 (RN , |µx |) ν − a.e. in x ∈ Ω, Z x 7→ h(x, z)dµx (z) ∈ L1 (Ω, ν) Ω

and E14

Z µZ

Z

(10.15)

¶ h(x, z)dµx (z) dν(x).

h(x, z)dµ(x, z) = Ω×RN

Ω

RN


e2

153

E14

From (10.14) and (10.15), we get, for ϕ ∈ D(Ω) and ψ ∈ D(RN ), Ã N Z ! ¶ Z µZ X ∂f ρ(z)zi ψ(z)dz ψ(z)dµx (z) ϕ(x)dν(x) = , ϕ(x) . ∂xi N Ω RN i=1 R Hence, as measures, N Z X i=1

RN

∂f ρ(z)zi ψ(z)dz = ∂xi

Z ψ(z)dµx (z) ν. RN

˜ Let now ψ˜ ∈ D(RN ) a radial function such that ψ˜ = 1 in supp(ρ). We set ψ(z) = ψ(z)z i. Then Z Z ∂f 2˜ ˜ ρ(z)zi ψ(z)dz = ψ(z)z i dµx (z) ν. ∂x N N i R R R E14 1 ˜ Since ν ∈ Mb (Ω) and x 7→ RN ψ(z)z i dµx (z) ∈ L (Ω, ν), f ∈ BV (Ω). Going back to (10.15) we obtain that N X ∂f µ(x, z) = (x) · ρ(z)zi LN (z). ∂x i i=1 BBM

bbm26

As in the proof of [23, Theorem 4], to prove 2 it is enough to show that for any δ > 0 there exists nδ ∈ N such that Z δ −N tN −1 Fn (t) dt ≤ Cδ q for n ≥ nδ (10.16) δ 0

for some constant C independent of n and δ, being Fn the function defined for t > 0 as Z Z Fn (t) = |fn ((x + tw) − fn (x)|q dx dσ N −1 N w∈S Z RZ 1 = N −1 |fn ((x + h) − fn (x)|q dx dσ. t |h|=t RN new:05

equat00

In terms of Fn assumption (10.11) can be expressed as Z 1 Fn (t) 1 (10.17) tN +q−1 q ρn (t) dt ≤ M q . t n 0 BBM

equat0000

On the other hand, applying [23, Lemma 2] with g(t) = Fn (t)/tq and h(t) = ρn (t), there exists a constant K = K(N + q) > 0 such that µZ δ ¶ µZ ¶ Z δ N +q−1 Fn (t) −N −q N +q−1 Fn (t) q t (10.18) δ dt ≤ K t ρn (t) |x| ρn (x)dx . tq tq 0 0 [|x| 0, we can find nδ ∈ N such that Z Z q |x| ρn (x) dx = |x|q nN ρ(nx) dx [|x| 0 and v ∈ W 1,p (Ω), and letting λ → 0, we get Z Z CJ,p J(z)χ(x, z)z · ∇v(x) dx dz RN Z Ω 2 Z CJ,p = J(z) |z · ∇u(x)|p−2 (z · ∇u(x)) (z · ∇v(x)) dx dz. 2 N R Ω Consequently, Z Z Z CJ,p J(z)χ(x, z)z · ∇v(x) dx dz = CJ,p a(∇u) · ∇v for every v ∈ W 1,p (Ω), 2 N R Ω Ω where Z 1 aj (ξ) = CJ,p J(z) |z · ξ|p−2 z · ξ zj dz. RN 2 Then, if we prove that E1AP

a(ξ) = |ξ|p−2 ξ,

(10.27) ecua:02

then (E1AP 10.25) is true and u = (I + Bp )−1 φ. So, to finish the proof we only need to show that (10.27) holds. Obviously, a is positively homogeneous of degree p − 1, that is, a(tξ) = tp−1 a(ξ) E1AP

for all ξ ∈ RN

and all t > 0.

Therefore, in order to prove (10.27) it is enough to see that ai (ξ) = ξi

for all ξ ∈ RN , |ξ| = 1

for each i.

t Now, let Rξ,i be the rotation such that Rξ,i (ξ) = ei , where ei is the vector with components t (ei )i = 1, (ei )j = 0 for j 6= i, and Rξ,i is the transpose of Rξ,i . Observe that t t t ξi = ξ · ei = Rξ,i (ξ) · Rξ,i (ei ) = ei · Rξ,i (ei ). R −1 On the other hand, since J is radial, CJ,p = 12 RN J(z)|zi |p dz and

a(ei ) = ei

for any i.

Then, if we make the change of variables z = Rξ,i (y), since J is a radial function, we obtain Z 1 ai (ξ) = CJ,p J(z)|z · ξ|p−2 z · ξ z · ei dz 2 N ZR 1 t = CJ,p J(y)|y · ei |p−2 y · ei y · Rξ,i (ei ) dy 2 N R t t = a(ei ) · Rξ,i (ei ) = ei · Rξ,i (ei ) = ξi . This ends the proof.

ecuacion03

¤

Theorem 134. Let Ω a smooth bounded domain in RN . Assume J(x) ≥ J(y) if |x| ≤ |y|. For any φ ∈ L∞ (Ω), ¡ ¢−1 (10.28) I + BpJp,ε φ → (I + Bp )−1 φ in Lp (Ω) as ε → 0.

158


new:02ecu:02

prop:01

Proof. The proof is a consequence of Proposition 133, (10.22), and Proposition 132. ¤

From the above theorem by standard results of the Nonlinear Semigroup Theory (see Cr2 BC BCP teo.3.intro.convergencia [50], [19] and [20]) we obtain the following result, which gives Theorem 130 in the case p > 1. conp>1

ecuacion02

Theorem 135. Let Ω be a smooth bounded domain in RN . Assume J(x) ≥ J(y) if J |x| ≤ |y|. Let T > 0 and u0 ∈ Lq (Ω), p ≤ q < +∞. Let uε the unique solution of Pp p,ε (u0 ) and u the unique solution of Np (u0 ). Then (10.29)

lim sup kuε (., t) − u(., t)kLq (Ω) = 0.

ε→0 t∈[0,T ]

ecuacion02

Moreover, if 1 < p ≤ 2, (10.29) holds for any u0 ∈ Lq (Ω), 1 ≤ q < +∞. ecuacion01

Proof. Since BpJ is completely accretive and satisfies the range condition (10.2), to ecuacion02 get (10.29) it is enough to see ¡ ¢−1 I + BpJp,ε φ → (I + Bp )−1 φ in Lq (Ω) as ε → 0 ³ ´−1 J for any φ ∈ L∞ (Ω). Taking into account that I + Bp p,ε φ ¿ φ, the above convergence ecuacion03 follows by (10.28). ¤ Convergence to the total variation flow for p = 1 As was mentioned in the introduction, motivated by problems in image processing, the problem N (u ), that is, the Neumann problem for the Total Variation Flow, was studied ABCM1 1 0 ACMBook in [2] (see also [3]). SolDebN

eaksolution

Definition 136. A measurable function u : (0, T )×Ω → R is a weak solution of N1 (u0 ) 1,1 in (0, T ) × Ω if u ∈ C([0, T ], L1 (Ω)) ∩ Wloc (0, T ; L1 (Ω)), Tk (u) ∈ L1w (0, T ; BV (Ω)) for all k > 0 and there exists z ∈ L∞ ((0, T ) × Ω) with kzk∞ ≤ 1, ut = div(z) in D0 ((0, T ) × Ω) such that Z Z (Tk (u(t)) − w)ut (t) dx ≤ z(t) · ∇w dx − |DTk (u(t))|(Ω) Ω 1,1

for every w ∈ W

Ω ∞

(Ω) ∩ L (Ω) and a.e. on [0, T ]. ABCM1

The main result of [2] is the following. Theorem 137. Let u0 ∈ L1 (Ω). Then there exists a unique weak solution of N1 (u0 ) in (0, T ) × Ω for every T > 0 such that u(0) = u0 . Moreover, if u(t), uˆ(t) are weak solutions corresponding to initial data u0 , uˆ0 , respectively, then k(u(t) − uˆ(t))+ k1 ≤ k(u0 − uˆ0 )+ k1 for all t ≥ 0.

and

ku(t) − uˆ(t)k1 ≤ ku0 − uˆ0 k1 ,


weaksolution

159

BC

Theorem 137 is proved using the techniques of completely accretive operators ([19]) and the Crandall-Liggett’s semigroup generation Theorem. To this end, the following operator ABCM1 B1 in L1 (Ω) was defined in [2] by the following rule: (u, v) ∈ B1

if and only if u, v ∈ L1 (Ω), Tk (u) ∈ BV (Ω) for all k > 0 and

there exists z ∈ L∞ (Ω, RN ) with kzk∞ ≤ 1, v = −div(z) in D0 (Ω) such that Z Z (w − Tk (u))v dx ≤ z · ∇wdx − |DTk (u)|(Ω), Ω

Ω

∀w ∈ W 1,1 (Ω) ∩ L∞ (Ω), ∀k > 0. weaksolution ABCM1 Theorem 137 follows from the following result given in [2].

cretivitatN

Theorem 138. The operator B1 is m-completely accretive in L1 (Ω) with dense domain. For any u0 ∈ L1 (Ω) the semigroup solution u(t) = e−tB1 u0 is a strong solution of   du + B u 3 0, 1 dt  u(0) = u0 . Set

1new:02

CJ,1 ³ x ´ , J1,ε (x) := 1+N J ε ε

with

1 1 := CJ,1 2

Z J(z)|zN | dz. RN

Theorem 139. Assume Ω is a smooth bounded domain in RN and J(x) ≥ J(y) if |x| ≤ |y|. For any φ ∈ L∞ (Ω), we have ³ ´−1 J I + B1 1,ε φ → (I + B1 )−1 φ in L1 (Ω) as ε → 0. ³ ´−1 J Proof. Given ε > 0, we set uε = I + B1 1,ε φ. Then, there exists g² ∈ L∞ (Ω × Ω), gε (x, y) = −gε (y, x) for almost all x, y ∈ Ω, kgε k∞ ≤ 1, µ ¶ µ ¶ x−y x−y J gε (x, y) ∈ J sign(uε (y) − uε (x)) a.e. x, y ∈ Ω ε ε and

ecu:01

(10.30) Observe that

===ecua:05

(10.31)

CJ,1 − 1+N ε

µ

Z J Ω

x−y ε µ

¶ gε (x, y)dy = φ(x) − uε (x)

a.e. x ∈ Ω.

¶ x−y J gε (x, y)dy uε (x) dx ε Ω Ω µ ¶ Z Z CJ,1 1 x−y = 1+N J |uε (y) − uε (x)| dy dx. ε 2 Ω Ω ε

CJ,1 − 1+N ε

Z Z

160


ecu:01

888ecu:01

By (10.30), we can write ¶ Z Z µ CJ,1 x−y J gε (x, y)(v(y) − v(x)) dxdy ε 2ε1+N Ω Ω µ ¶ Z Z CJ,1 x−y (10.32) = − 1+N J gε (x, y)dyv(x) dx ε Ω Ω Zε =

∀v ∈ L∞ (Ω).

(φ(x) − uε (x))v(x) dx, Ω

Since uε ¿ φ, there exists a sequence εn → 0 such that uεn * u weakly in L1 (Ω),

u ¿ φ.

Observe that kuεn kL∞ (Ω) , kukL∞ (Ω) ≤ kφkL∞ (Ω) . ===ecua:05 Hence taking ε = εn and v = uεn in 888ecu:01 (10.32), changing variables and having in mind (10.31), we get ¯ ¯ Z Z ¯ uεn (x + εn z) − uεn (x) ¯ CJ,1 ¯ dx dz J(z)χΩ (x + εn z) ¯¯ ¯ ε n RN Ω 2 ¯ µ ¶¯ Z Z x − y ¯¯ uεn (y) − uεn (x) ¯¯ 1 CJ,1 = J ¯ ¯ dx dy N 2 ε ε ε n n n Ω Ω Z = (φ(x) − uεn (x))uεn (x) dx ≤ M, ∀ n ∈ N. Ω

prop:01

Therefore, by Proposition 132, u ∈ BV (Ω), 77ecua:03

(10.33)

CJ,1 uε (x + εn z) − uεn (x) CJ,1 J(z)χΩ (x + εn z) n * J(z)z · Du 2 εn 2

weakly as measures and uεn → u,

strongly in L1 (Ω).

Moreover, we also can assume that EDDD

ED9876

J(z)χΩ (x + εn z)gεn (x, x + εn z) * Λ(x, z)

(10.34)

weakly∗ in L∞ (Ω) × L∞ (RN ), and Λ(x, z) ≤ J(z) almost every where in Ω × RN . Changing 888ecu:01 variables and having in mind (10.32), we can write Z Z CJ,1 v(x + εn z) − v(x) J(z)χΩ (x + εn z)gεn (x, x + εn z) dz dx 2 RN ZΩ Z εn CJ,1 (10.35) =− J(z)χΩ (x + εn z)gεn (x, x + εn z) dz v(x) dx ε N Ω n R Z = (φ(x) − uεn (x))v(x) dx ∀v ∈ L∞ (Ω). Ω


EDDD

161

ED9876

By (10.34), passing to the limit in (10.35), we get

ED999

CJ,1 2Z

(10.36)

Z

Z Λ(x, z)z · ∇v(x) dx dz RN

=

Ω

∀v ∈ L∞ (Ω) ∩ W 1,1 (Ω).

(φ(x) − u(x))v(x) dx Ω

We set ζ = (ζ1 , . . . , ζN ), the vector field defined by CJ,1 ζi (x) := 2

Z Λ(x, z)zi dz,

i = 1, . . . , N.

RN

ED999

Then, ζ ∈ L∞ (Ω, RN ), and from (10.36), −div(ζ) = φ − u

in D0 (Ω).

Let us see that kζk∞ ≤ 1. Given ξ ∈ RN \ {0}, let Rξ be the rotation such that Rξt (ξ) = e1 |ξ|. Then, if we make the change of variables z = Rξ (y), we obtain CJ,1 ζ(x) · ξ = 2

Z

CJ,1 Λ(x, z)z · ξ dz = 2 RN CJ,1 = 2

Z Λ(x, Rξ (y))Rξ (y) · ξ dy RN

Z Λ(x, Rξ (y))y1 |ξ| dy. RN

On the other hand, since J is a radial function and Λ(x, z) ≤ J(z) almost every where, CJ,1 and CJ,1 |ζ(x) · ξ| ≤ 2

−1

1 = 2

Z J(z)|z1 | dz RN

Z J(y)|y1 | dy|ξ| = |ξ|

a.e. x ∈ Ω.

RN

Therefore, we obtain that kζk∞ ≤ 1. Since u ∈ L∞ (Ω), to finish the proof we only need to show that Z e12222

Z

(10.37)

(w − u)(φ − u) dx ≤ Ω

ζ · ∇wdx − |Du|(Ω), Ω

∀w ∈ W 1,1 (Ω) ∩ L∞ (Ω).

162


ED9876

e1000

Given w ∈ W 1,1 (Ω) ∩ L∞ (Ω), taking v = w − uεn in (10.35), we get Z (φ(x) − uεn (x))(w(x) − uεn (x)) dx Ω Z Z CJ,1 = J(z)χΩ (x + εn z)gεn (x, x + εn z) dz× 2 RN Ω µ ¶ w(x + εn z) − w(x) uεn (x + εn z) − uεn (x) × − dx (10.38) εn εn Z Z CJ,1 w(x + εn z) − w(x) dx = J(z)χΩ (x + εn z)gεn (x, x + εn z) dz 2 RN Ω εn ¯ ¯ Z Z ¯ uε (x + εn z) − uεn (x) ¯ CJ,1 ¯ dx. − J(z)χΩ (x + εn z) ¯¯ n ¯ 2 RN Ω εn 77ecua:03

EDDD

e1000

Having in mind (10.33) and (10.34) and taking limit in (10.38) as n → ∞, we obtain that Z Z Z Z Z CJ,1 CJ,1 (w − u)(φ − u) dx ≤ Λ(x, z)z · ∇w(x) dx dz − |J(z)z · Du| 2 Ω RN 2 Ω RN Ω Z Z Z CJ,1 = ζ · ∇w dx − |J(z)z · Du|. 2 Ω RN Ω Du Now, for every x ∈ Ω such that the Radon-Nikodym derivative |Du| (x) 6= 0, let Rx be Du t Du the rotation such that Rx [ |Du| (x)] = e1 | |Du| (x)|. Then, since J is a radial function and Du | |Du| (x)| = 1 |Du|-a.e. in Ω, if we make the change of variables y = Rx (z), we obtain ¯ ¯ Z Z Z Z ¯ ¯ CJ,1 CJ,1 Du (x)¯¯ dz d|Du|(x) |J(z)z · Du| = J(z) ¯¯z · 2 Ω RN 2 Ω RN |Du|

CJ,1 = 2 e12222

Z Z

Z J(y) |y1 | dy d|Du|(x) =

Ω RN

|Du|. Ω

Consequently, (10.37) holds and the proof concludes.

¤ conp>1

From the above theorem, arguing as in Theorem 135, by standard results of the NonCr2 BCP linear Semigroup Theory ( [50], [20]), we obtain the following result, from which Theorem teo.3.intro.convergencia 130 holds in the case p = 1. Theorem 140. Let Ω a smooth bounded domain in RN . Assume J(x) ≥ J(y) if |x| ≤ |y|. Let T > 0 and u0 ∈ L1 (Ω). Let uε the unique solution in [0, T ] of P1J (u0 ) and u the unique weak solution of N1 (u0 ). Then lim sup kuε (., t) − u(., t)kL1 (Ω) = 0.

ε→0 t∈[0,T ]


163

0.29. Asymptotic behaviour. Now we study the asymptotic behaviour as t → ∞ of the solutions of the nonlocal problems. We show that the solutions of the nonlocal problems converge to the mean value of the initial condition. teo.asymp

Theorem 141. Let p ≥ 1 and u0 ∈ L∞ (Ω). Let u be the solution to PpJ (u0 ), then Ã !1/p ||u0 ||2L2 (Ω) ku(t) − u0 kLp (Ω) ≤ → 0, as t → ∞, t where u0 is the mean value of the initial condition, Z 1 u0 = u0 (x) dx. |Ω| Ω teo.asymp

To prove Theorem 141, we start by showing the following Poincaré’s type inequality. In the linear ChChR case, that is, for p = 2, Poincaré’s type inequality has been proved using spectral theory in [34]. blg:01

Proposition 142. Given p ≥ 1, J and Ω, the quantity Z Z 1 J(x − y)|u(y) − u(x)|p dy dx 2 Ω Ω Z βp−1 := βp−1 (J, Ω, p) = infR u∈Lp (Ω), Ω u=0 |u(x)|p dx Ω

.c.Poincare

001

is strictly positive. Consequently Z ¯ Z ¯p Z Z ¯ ¯ 1 1 ¯ ¯ (10.39) βp−1 ¯u − u¯ ≤ J(x − y)|u(y) − u(x)|p dy dx |Ω| 2 Ω Ω Ω Ω

Proof. It is enough to prove that there exists a constant c such that ÃµZ Z ¶1/p ¯Z ¯! ¯ ¯ (10.40) kukp ≤ c J(x − y)|u(y) − u(x)|p dydx + ¯¯ u¯¯ ∀u ∈ Lp (Ω). Ω

008

∀u ∈ Lp (Ω).

Ω

Ω

Let r > 0 such that J(z) ≥ α > 0 in B(0, r). Since Ω ⊂ ∪x∈Ω B(x, r/2), there exists m {xi }m i=1 ⊂ Ω such that Ω ⊂ ∪i=1 B(xi , r/2). Let 0 < δ < r/2 such that B(xi , δ) ⊂ Ω for all i = 1, ...m. Then, for any xî ∈ B(xi , δ), i = 1, ..., m, m [ (10.41) Ω = (B(ˆ xi , r) ∩ Ω). i=1

001

Let us argue by contradiction. Suppose that (10.40) is false. Then, there exists un ∈ p L (Ω), kun kp = 1, and satisfying ÃµZ Z ¯! ¶1/p ¯Z ¯ ¯ ∀n ∈ N. + ¯¯ un ¯¯ 1≥n J(x − y)|un (y) − un (x)|p dydx Ω

Ω

Ω

164


Consequently, 003

(10.42)

Z Z J(x − y)|un (y) − un (x)|p dydx = 0

lim n

Ω

Ω

and 004

Z

(10.43)

lim n

un = 0. Ω

Let Fn (x, y) = J(x − y)1/p |un (y) − un (x)| and

Z J(x − y)|un (y) − un (x)|p dy.

fn (x) = Ω

003

From (10.42), it follows that fn → 0 in L1 (Ω), so we can assume there exists a subsequence, denoted equal, such that 005

(10.44)

fn (x) → 0 ∀x ∈ Ω \ B1 ,

003

B1 null.

On the other hand, by (10.42), we also have that Fn → 0 en Lp (Ω × Ω). So we can suppose, passing to a subsequence if necessary, 006

(10.45)

Fn (x, y) → 0 ∀(x, y) ∈ Ω × Ω \ C,

C null.

Let B2 ⊂ Ω a null set satisfying that, 007

for all x ∈ Ω \ B2 , the chapter Cx of C is null.

(10.46)

Let xˆ1 ∈ B(x1 , δ) \ (B1 ∪ B2 ), then there exists a subsequence, denoted equal, such that un (ˆ x1 ) → λ1 ∈ [−∞, +∞]. Consider now xˆ2 ∈ B(x2 , δ) \ (B1 ∪ B2 ), then up to a subsequence, we can assume un (ˆ x2 ) → λ2 ∈ [−∞, +∞]. So, successively (up to m), for xˆm ∈ B(xm , δ) \ (B1 ∪ B2 ), there exists a subsequence, againdenoted equal, such that un (ˆ xm ) → λm ∈ [−∞, +∞]. 006

007

By (10.45) and (10.46), 008

un (y) → λi

∀y ∈ (B(ˆ xi , r) ∩ Ω) \ Cxî .

Now, by (10.41), Ω = (B(ˆ x1 , r) ∩ Ω) ∪ (∪m xi , r) ∩ Ω)). i=2 (B(ˆ


165

Hence, since Ω is a domain, there exists i2 ∈ {2, .., m} such that (B(ˆ x1 , r) ∩ Ω) ∩ (B(ˆ xi2 , r) ∩ Ω) 6= ∅. Therefore, λ1 = λi2 . Let us call i1 := 1. Again, since xi , r) ∩ Ω)), Ω = ((B(ˆ xi1 , r) ∩ Ω) ∪ ((B(ˆ xi1 , r) ∩ Ω)) ∪ (∪i∈{1,...,m}\{i1 ,i2 } (B(ˆ and there exists i3 ∈ {1, ..., m} \ {i1 , i2 } such that ((B(ˆ xi1 , r) ∩ Ω) ∪ ((B(ˆ xi1 , r) ∩ Ω)) ∩ (B(ˆ xi3 , r) ∩ Ω) 6= ∅. Consequently λi1 = λi2 = λi3 . Using the same argument we arrive at λ1 = λ2 = ... = λm = λ. If |λ| = +∞, we have shown that |un (y)|p → +∞ for almost every y ∈ Ω, which contradicts kun kp = 1 for all n ∈ N. Hence λ is finite. 005

On the other hand, by (10.44), fn (ˆ xi ) → 0, i = 1, ..., m, hence, Fn (ˆ x1 , .) → 0 in Lp (Ω). Since un (ˆ x1 ) → λ, from the above we conclude that un → λ in Lp (B(ˆ xi , r) ∩ Ω). Using again the compactness argument we get un → λ in Lp (Ω).

004

Now, by (10.43), λ = 0, so un → 0 in Lp (Ω), which contradicts kun kp = 1.

¤

Remark 143. The above Poincaré’s type inequality fails to be true in general if 0 ∈ / supp(J), as the following example shows. Let Ω = (0, 3) and J be such that supp(J) ⊂ (−3, −2) ∪ (2, 3). Then, if

( u(x) =

we have that

Z

3 0

Z

3 0

1

if 0 < x < 1 or 2 < x < 3,

2

1 ≤ x ≤ 2,

J(x − y)|u(y) − u(x)|p dx dy = 0,

166


but clearly

Z 1 3 u(x) − u(y) dy 6= 0. 3 0 Therefore there is no Poincaré’s type inequality available for this J. This example can be easily extended for any domain in any dimension just by considering functions u that are constant on an annuli intersected with Ω. teo.asymp

Next we prove Theorem 141.

teo.asymp

Proof of Theorem 141. First we observe that a simple integration in space of the equation gives that the total mass is preserved, that is, Z Z 1 1 u(x, t) dx = u0 (x) dx. |Ω| Ω |Ω| Ω Let

1 w(x, t) = u(x, t) − |Ω|

Z u0 (x) dx. Ω

Then, Z Z Z d p p−2 |w(x, t)| dx = p |w| w(x, t) J(x−y)|w(y, t)−w(x, t)|p−2 (w(y, t)−w(x, t)) dydx dt Ω Ω Ω Z Z p J(x−y)|w(y, t)−w(x, t)|p−2 (w(y, t)−w(x, t))(|w|p−2 w(y, t)−|w|p−2 w(x, t)) dydx. =− 2 Ω Ω Therefore the Lp (Ω)-norm of w is decreasing with t. Moreover, as the solution preserves the total mass, using Poincaré’s type inequality Imejor.c.Poincare (10.39), we have, Z Z Z p |w(x, t)| dx ≤ C J(x − y)|u(y, t) − u(x, t)|p dy dx. Ω

Ω

Ω

Consequently, Z Z tZ Z tZ Z p p t |w(x, t)| dx ≤ |w(x, s)| dx ds ≤ C J(x − y)|u(y, s) − u(x, s)|p dy dx ds. Ω

0

Ω

0

Ω

Ω

On the other hand, multiplying the equation by u(x, t) and integrating in space and time, we get Z Z Z tZ Z 2 2 |u(x, t)| − |u0 (x)| dx = − J(x − y)|u(y, s) − u(x, s)|p dy dx ds, Ω

which implies

Ω

0

Z tZ Z 0

Ω

Ω

Ω

Ω

J(x − y)|u(y, s) − u(x, s)|p dy dx ds ≤ ||u0 ||2L2 (Ω) ,


and therefore

Z |w(x, t)|p dx ≤

||u0 ||2L2 (Ω) t

Ω

167

. ¤ Imejor.c.Poincare

Remark 144. Observe that using Poincaré’s type inequality (10.39), we can obtain mazon

u + BpJ u = φ,

(10.47)

for p ≥ 2 in the following manner: let ½ ¾ Z p K := u ∈ L (Ω) : u=0 Ω p0

and A : K → L (Ω) the continuous monotone operator defined by A(u) := u + BpJ u. By Imejor.c.Poincare (10.39), we have Z A(u)u Ω lim = +∞. kukp kukp → +∞ u∈K R Br1 Then, by Corollary 30 in [26], for φ ∈ L∞ (Ω), Ω φ = 0, there exists u ∈ K, such that Z Z Z J uv + Bp uv = φv ∀v ∈ K. Ω Ω Ω R R R Since Ω u = 0, Ω φ = 0 and Ω BpJ u = 0, we have that µ Z ¶ Z Z ¶ Z Z Z µ 1 1 J J v + Bp u v − v uv + Bp uv = u v− |Ω| Ω |Ω| Ω Ω Ω ZΩ µ Z ¶ ZΩ 1 = φ v− v = φv, |Ω| Ω Ω Ω mazon

for any v ∈ Lp (Ω), and consequently (10.47) holds.

CHAPTER 11

The limit as p → ∞ in a nonlocal p−Laplacian evolution equation. A nonlocal approximation of a model for sandpiles

c.sandpiles

Our main purpose in this chapter is to study a nonlocal ∞−Laplacian type diffusion equation obtained as the limit as p → ∞ to the nonlocal analogous to the p−Laplacian evolution. EFG

AEW

First, let us recall some known results on local evolution problems. In [56] (see also [10] Evans and [55]) was investigated the limiting behavior as p → ∞ of solutions to the quasilinear parabolic problem   vp,t − ∆p vp = f, in ]0, T [×RN , Pp (u0 )  v (0, x) = u (x), in RN , p 0 where ∆p u = div (|∇u|p−2 ∇u) and f is nonnegative and represents a given source term, which is interpreted physically as adding material to an evolving system, within which mass particles are continually rearranged by diffusion. We hereafter take the space H = L2 (RN ) and define for 1 < p < ∞ the functional  Z  1 if u ∈ L2 (RN ) ∩ W 1,p (RN ), |∇v(y)|p dy, p RN Fp (v) =  +∞ if u ∈ L2 (RN ) \ W 1,p (RN ). Therefore, the PDE problem Pp (u0 ) has the standard reinterpretation   f (t) − vp,t = ∂Fp (vp (t)), a.e. t ∈]0, T [,  v (0, x) = u (x), p 0

in RN .

EFG

In [56], assuming that u0 is a Lipschitz function with compact support, satisfying k∇u0 k∞ ≤ 1, and for f a smooth nonnegative function with compact support in [0, T ] × RN , it is proved that we can extract a sequence pi → +∞ and obtain a limit function v∞ , such that for each T > 0,   vpi → v∞ , a.e. and in L2 (RN × (0, T )),  ∇v * ∇v , v * v pi ∞ pi ,t ∞,t

weakly in L2 (RN × (0, T )). 169

170

11. A MODEL FOR SANDPILES

Moreover, the limit function v∞ satisfies   f (t) − v∞,t ∈ ∂F∞ (v∞ (t)), P∞ (u0 )  v (0, x) = u (x), ∞ 0 where F∞ (v) =

a.e. t ∈]0, T [, in RN ,

  0

if v ∈ L2 (RN ), |∇v| ≤ 1,

 +∞

in other case.

This limit problem P∞ (u0 ) explains the movement of a sandpile (v∞ (x, t) describes the amount of the sand at the point x at time t), the main assumption being that the sandpile is stable when the slope is less or equal than one and unstable if not. On the other hand, we have the following nonlocal nonlinear diffusion problem, which we call the nonlocal p-Laplacian problem, Z   up,t (x, t) = J(x − y)|up (y, t) − up (x, t)|p−2 (up (y, t) − up (x, t))dy + f (x, t), J N Pp (u0 ) R  up (0, x) = u0 (x). Here J : RN R→ R is a nonnegative continuous radial function with compact support, J(0) > 0 and RN J(x) dx = 1 (this last condition is not necessary to prove our results, it is imposed to simplify the exposition). AMRTp-Lap In the previous chapter (see aso [9]) we have studied this problem when the integral is taken in a bounded domain Ω (hence dealing with homogeneous Neumann boundary conditions). We have obtained existence and uniqueness of solutions and, if the kernel J is rescaled in an appropriate way, that the solutions to the corresponding nonlocal problems converge to the solution of the p−laplacian with homogeneous Neumann boundary conditions. We have also studied the asymptotic behaviour of the solutions as t goes to infinity, showing the convergence to the mean value of the initial condition. Let us note that the evolution problem PpJ (u0 ) is the gradient flow associated to the functional Z Z 1 J Gp (u) = J(x − y)|u(y) − u(x)|p dy dx, 2p RN RN which is the nonlocal analogous to the functional Fp associated to the p−Laplacian. AMRTp-Lap

Following [9], we obtain existence and uniqueness of a global solution for this nonlocal problem. Our next result in this article concerns the limit as p → ∞ in PpJ (u0 ). We obtain that the limit functional is given by   0 if u ∈ L2 (RN ), |u(x) − u(y)| ≤ 1, for x − y ∈ supp(J), J G∞ (u) =  +∞ in other case.

cia.p.intro


Then, the nonlocal limit problem can be written as   f (t, ·) − ut (t, ·) ∈ ∂GJ∞ (u(t)), J P∞ (u0 )  u(0, x) = u (x). 0

171

a.e. t ∈]0, T [,

With these notations, we obtain the following result. Theorem 145. Let T > 0, f ∈ BV (0, T ; Lp (RN )), u0 ∈ L2 (RN ) ∩ Lp (RN ) such that |u0 (x) − u0 (y)| ≤ 1, for x − y ∈ supp(J), and up the unique solution of PpJ (u0 ). Then, if J u∞ is the unique solution to P∞ (u0 ), lim sup kup (t, ·) − u∞ (t, ·)kL2 (RN ) = 0.

p→∞ t∈[0,T ]

) we 0.30. Preliminaries. To identify the limit of the solutions up of problem PpJ (u0ET will use the methods of convex analysis, and so we first recall some terminology (see [53], Brezis Attouch [27] and [11]). If H is a real Hilbert space with inner product ( , ) and Ψ : H → (−∞, +∞] is convex, then the subdifferential of Ψ is defined as the multivalued operator ∂Ψ given by v ∈ ∂Ψ(u) ⇐⇒ Ψ(w) − Ψ(u) ≥ (v, w − u) ∀ w ∈ H. The epigraph of Ψ is defined by Epi(Ψ) = {(u, λ) ∈ H × R : λ ≥ Ψ(u)}. Given K a closed convex subset of H, the indicator function of K is defined by   0 if u ∈ K, IK (u) =  +∞ if u 6∈ K. Then it is easy to see that the subdifferential is characterized as follows, v ∈ ∂IK (u) ⇐⇒ u ∈ K and (v, w − u) ≤ 0 ∀ w ∈ K. In case the convex functional ΨBrezis : H → (−∞, +∞] is proper , lower-semicontinuous and min Ψ = 0 , it is well known (see [27]) that the abstract Cauchy problem   u0 (t) + ∂Ψ(u(t)) 3 f (t), a.e t ∈]0, T [,  u(0) = u , 0 has a unique strong solution for any f ∈ L2 (0, T ; H) and u0 ∈ D(∂Ψ). Mosco

Attouch

The following convergence was studied by Mosco in [72] (see [11]). Suppose X is a metric space and An ⊂ X. We define lim inf An = {x ∈ X : ∃xn ∈ An , xn → x} n→∞

172


and lim sup An = {x ∈ X : ∃xnk ∈ Ank , xnk → x}. n→∞

In the case X is a normed space, we note by s − lim and w − lim the above limits associated respectively to the strong and to the weak topology of X. Given a sequence Ψn , Ψ : H → (−∞, +∞] of convex lower-semicontinuous functionals, we say that Ψn converges to Ψ in the sense of Mosco if e1Mos

(11.1)

w − lim sup Epi(Ψn ) ⊂ Epi(Ψ) ⊂ s − lim inf Epi(Ψn ). n→∞

n→∞

e1Mos

It is easy to see that (11.1) is equivalent to the two following conditions: e2Mos

(11.2)

∀ u ∈ D(Ψ) ∃un ∈ D(Ψn ) : un → u and Ψ(u) ≥ lim sup Ψn (un ); n→∞

e3Mos

ncia1.Mosco

(11.3)

for every subsequence nk , when uk * u, it holds Ψ(u) ≤ lim inf Ψnk (uk ). k

BP

Attouch

As consequence of results in [28] and [11] we can write the following result. Theorem 146. Let Ψn , Ψ : H → (−∞, +∞] convex lower-semicontinuous functionals. Then the following statements are equivalent: (i) Ψn converges to Ψ in the sense of Mosco. (ii) (I + λ∂Ψn )−1 u → (I + λ∂Ψ)−1 u,

∀ λ > 0, u ∈ H.

Moreover, any of these two conditions (i) or (ii) imply that (iii) for every u0 ∈ D(∂Ψ) and u0,n ∈ D(∂Ψn ) such that u0,n → u0 , and every fn , f ∈ L1 (0, T ; H) with fn → f , if un (t), u(t) are the strong solutions of the abstract Cauchy problems   u0n (t) + ∂Ψn (un (t)) 3 fn , a.e. t ∈]0, T [,  u (0) = u , n 0,n and

  u0 (t) + ∂Ψ(u(t)) 3 f,

a.e. t ∈]0, T [,

 u(0) = u , 0 respectively, then un → u

in C([0, T ] : H).

Let us also collect some preliminaries and notations concerning completely accretive BC operators that will be used afterwards (see [19]). We denote by J0 and P0 the following sets of functions, J0 = {j : R → [0, +∞], such that j is convex, lower semi-continuos and j(0) = 0},


173

P0 = {q ∈ C ∞ (R) : 0 ≤ q 0 ≤ 1, supp(q 0 ) is compact, and 0 ∈ / supp(q)} . Let M (RN ) denote the space of measurable functions from RN into R. We set L(RN ) := L (RN ) + L∞ (RN ). Note that L(RN ) is a Banach space with the norm 1

kuk1+∞ := inf{kf k1 + kgk∞ : f ∈ L1 (RN ), g ∈ L∞ (RN ), f + g = u}. The closure of L1 (RN ) ∩ L∞ (RN ) in L(RN ) is denoted by L0 (RN ). Given u, v ∈ M (RN ) we say

Z

Z

j(v) dx

j(u) dx ≤

u ¿ v if and only if RN

∀j ∈ J0 .

RN

An operator A ⊂ M (RN ) × M (RN ) is said to be completely accretive if u − uˆ ¿ u − uˆ + λ(v − vˆ)

∀λ > 0 and ∀(u, v), (ˆ u, vˆ) ∈ A.

BC

The following facts are proved in [19]. 00.2

ef.weak.def

Proposition 147. (i) Let u ∈ L0 (RN ), v ∈ L(RN ), then u ¿ u + λv

Z

∀λ > 0 if and only if

q(u)v ≥ 0,

∀q ∈ P0 .

RN

(ii) If v ∈ L0 (RN ), then {u ∈ M (RN ) : u ¿ v} is a weak sequentially compact subset of L0 (RN ). AMRTp-Lap

Concerning nonlocal models, in [9] we have studied the following nonlocal nonlinear diffusion problem, which we call the nonlocal p-Laplacian problem with homogeneous Neumann boundary conditions, Z   ut (x, t) = J(x − y)|u(y, t) − u(x, t)|p−2 (u(y, t) − u(x, t)) dy, Ω  u(0, x) = u0 (x). Using similar ideas and techniques we can deal with the nonlocal problem in RN . Solutions to PpJ (u0 ) are to be understood in the following sense. Definition 148. Let 1 < p < +∞. Let f ∈ L1 (0, T ; Lp (RN )) and u0 ∈ Lp (RN ). A solution of PpJ (u0 ) in [0, T ] is a function u ∈ C([0, T ]; Lp (RN ))∩W 1,1 (]0, T [; Lp (RN )) which satisfies u(0, x) = u0 (x) a.e. x ∈ RN and Z ut (x, t) = J(x−y)|u(y, t)−u(x, t)|p−2 (u(y, t)−u(x, t)) dy+f (x, t) a.e. in ]0, T [×RN . RN

174


AMRTp-Lap

Working as in [9], we can obtain the following result about existence and uniqueness 0 of a global solution for this problem. Let us first define BpJ : Lp (RN ) → Lp (RN ) by Z J J(x − y)|u(y) − u(x)|p−2 (u(y) − u(x)) dy, x ∈ RN . Bp u(x) = − RN

DESIGU1.2

o.1.intro.2

cuacion01.2

Observe that, for every u, v ∈ Lp (RN ) and T : R → R such that T (u − v) ∈ Lp (RN ), it holds Z (BpJ u(x) − BpJ v(x))T (u(x) − v(x))dx = N R Z Z 1 (11.4) J(x − y) (T (u(y) − v(y)) − T (u(x) − v(x))) × 2 RN RN ¡ ¢ × |u(y) − u(x)|p−2 (u(y) − u(x)) − |v(y) − v(x)|p−2 (v(y) − v(x)) dy dx. Let us also define the operator © ª BpJ = (u, v) ∈ Lp (RN ) × Lp (RN ) : v = BpJ (u) . It is easy to see that Dom(BpJ ) = Lp (RN ) and BpJ is positively homogeneous of degree p − 1. Theorem 149. Let 1 < p < +∞. If f ∈ BV (0, T ; Lp (RN )) and u0 ∈ D(BpJ ) then there exists a unique solution to PpJ (u0 ). If f = 0 then there exists a unique solution to PpJ (u0 ) for all u0 ∈ Lp (RN ). Moreover, if ui (t) is a solution of PpJ (ui 0 ) with f = fi , fi ∈ L1 (0, T ; Lp (RN )) and ui 0 ∈ Lp (RN ), i = 1, 2, then, for every t ∈ [0, T ], Z t + + kf1 (s) − f2 (s)kLp (RN ) ds. k(u1 (t) − u2 (t)) kLp (RN ) ≤ k(u1 0 − u2 0 ) kLp (RN ) + 0

Proof. Let us first show that BpJ is completely accretive and verifies the following range condition Lp (RN ) = Ran(I + BpJ ).

(11.5)

DESIGU1.2

Indeed, given ui ∈ Dom(BpJ ), i = 1, 2 and q ∈ P0 , by (11.4) we have Z (BpJ u1 (x) − BpJ u2 (x)) q(u1 (x) − u2 (x)) dx ≥ 0, RN

from where it follows that BpJ is a completely accretive operator. To show that BpJ satisfies the range condition we have to prove that for any φ ∈ Lp (RN ) there exists u ∈ Dom(BpJ ) such that u = (I + BpJ )−1 φ. Let us first take φ ∈ L1 (RN ) ∩ L∞ (RN ). For every n ∈ N, let AMRTp-Lap J defined by φn := φχBn (0) . By the results in [9], the operator Bp,n Z J u(x) = − J(x − y)|u(y) − u(x)|p−2 (u(y) − u(x)) dy, x ∈ Bn (0), Bp,n Bn (0)


175

is m-completely accretive in Lp (Bn (0)). Then, there exists un ∈ Lp (Bn (0)), such that e1Mayte

J un (x) + Bp,n un (x) = φn (x),

(11.6)

a.e. in Bn (0).

Moreover, un ¿ φn . We denote by uñ and Hn the extensions   un (x) if x ∈ Bn (0), uñ (x) =  0 if x ∈ RN \ Bn (0), and Hn (x) =

 J  Bp,n un (x), if x ∈ Bn (0),  0

if x ∈ RN \ Bn (0).

Since, un ¿ φn , we have e2Mayte

(11.7)

k˜ un kq ≤ kφkq

for all 1 ≤ q ≤ ∞, ∀ n ∈ N.

Hence, we can suppose 0

uñ * u in Lp (RN ).

e5Mayte

(11.8)

e3Mayte

On the other hand, multiplying (11.6) by un and integrating, we get Z Z (11.9) J(x − y)|un (y) − un (x)|p dydx ≤ kφk2 ∀ n ∈ N,

e1Mayte

Bn (0)

Bn (0) 0

which implies, by Hölder’s inequality, that {Hn : n ∈ N} is bounded in Lp (RN ). Therefore, we can assume that e4Mayte

(11.10)

e5Mayte

Hn * H

e4Mayte

0

in Lp (RN ).

e1Mayte

By (11.8) and (11.10), taking limit in (11.6), we get e7Mayte

(11.11) Let us see that

marfon4

(11.12)

a.e. in RN .

u+H =φ Z

J(x − y)|u(y) − u(x)|p−2 (u(y) − u(x)) dy

H(x) = −

a.e. in x ∈ RN .

RN

e1Mayte

In fact, multiplying (11.6) by un and integrating, we get Z Z J (φ − un )un Bp,n un un = B(0,n) B(0,n) Z Z Z Z = (φ − u)u − φ(u − un ) + 2u(u − un ) − B(0,n)

e7Mayte

B(0,n)

Therefore, by (11.11), marfon1

(11.13)

B(0,n)

Z

Z

lim sup Bn (0)

J un Bp,n

(u − un )(u − un ). B(0,n)

Z

un ≤

(φ − u)u = RN

H u. RN

176


On the other hand, for any v ∈ L1 (RN ) ∩ L∞ (RN ), since Z ¢ ¡ J J 0≤ Bp,n un − Bp,n v (un − v), Bn (0)

we have that, Z

Z

Bn (0)

J Bp,n un

un +

marfon1

Therefore, by (11.13), marfon2

Z Bn (0)

Z

J Bp,n v

v≥ Bn (0)

Z

(11.14)

Z

Hu+ RN

Z J Bp,n un

RN

BpJ v

v+ Bn (0)

J Bp,n v un .

Z

v≥

Hv+ RN

RN

BpJ v u.

Taking now v = u ± λw, λ > 0 and w ∈ L1 (RN ) ∩ L∞ (RN ), and letting λ → 0, we get Z Z Hw= BpJ u w, RN

marfon4

RN

e7Mayte

and consequently (11.12) is proved. Therefore, by (11.11), the range condition is satisfied for φ ∈ L1 (RN ) ∩ L∞ (RN ). Let now φ ∈ Lp (RN ). Take φn ∈ L1 (RN ) ∩ L∞ (RN ), φn → φ in Lp (RN ). Then, by our previous step, there exists un = (I + BpJ )−1 φn . Since BpJ is completely accretive, un → u in 0 Lp (RN ), also BpJ un → BpJ u in Lp (RN ) and we conclude that u + BpJ u = φ. BC

font1

BCP

Consequently (see [19] and [20]) we have that BpJ is an m-accretive operator in Lp (RN ) and we get the existence of mild solution u(t) of the abstract Cauchy problem   u0 (t) + BpJ u(t) = f (t), t ∈]0, T [, (11.15)  u(0) = u , 0 BCP

Cr

Cr2

Now, by the Nonlinear Semigroup Theory (see font1 [20], [49] or [50]), if f ∈ BV (0, T ; Lp (RN )) J and u0 ∈ D(Bp ), u(t) is a strong solution of (11.15), that is, a solution of PpJ (u0 ) in the def.weak.def sense of Definition 148. The same is true for all u0 ∈ Lp (RN ) in the case f = 0 by the complete accretivity of BpJ , since BC Dom(B2J ) = L2 (RN ) and for p 6= 2 the operator BpJ is homogeneous of degree p − 1 (see [19]). Finally, the contraction principle follows from the general Nonlinear Semigroup Theory since the solutions ui , i = 1, 2, are mild-solutions of font1 ¤ (11.15). 0.31. Limit as p → ∞. Recall from the Introduction that the nonlocal p-Laplacian evolution problem Z   ut (x, t) = J(x − y)|u(y, t) − u(x, t)|p−2 (u(y, t) − u(x, t)) dy + f (x, t), J Pp (u0 ) RN  u(0, x) = u0 (x).


177

is the gradient flow associated to the functional Z Z 1 J Gp (u) = J(x − y)|u(y) − u(x)|p dy dx. 2p RN RN With a formal calculation, taking limit as p → ∞, we arrive to the functional   0 if |u(x) − u(y)| ≤ 1, for x − y ∈ supp(J), J G∞ (u) =  +∞ in other case. Hence, if we define J := {u ∈ L2 (RN ) : |u(x) − u(y)| ≤ 1, for x − y ∈ supp(J)}, K∞ J J . we have that the functional GJ∞ is given by the indicator function of K∞ , that is, GJ∞ = IK∞ Then, the nonlocal limit problem can be written as   f (t, ·) − ut (t) ∈ ∂IK∞ J (u(t)), a.e. t ∈]0, T [, J P∞ (u0 )  u(0, x) = u (x). 0

convergencia.p.intro

Proof of Theorem 145. Let T > 0. Recall that we want to prove that, given f ∈ J BV (0, T ; Lp (RN )), u0 ∈ K∞ ∩ Lp (RN ) and up the unique solution of PpJ (u0 ), if u∞ is the J unique solution of P∞ (u0 ), then lim sup kup (t, ·) − u∞ (t, ·)kL2 (RN ) = 0.

p→∞ t∈[0,T ]

convergencia1.Mosco

By Theorem 146, to prove the result it is enough to show that the functionals Z Z 1 J Gp (u) = J(x − y)|u(y) − u(x)|p dy dx 2p RN RN converge to GJ∞ (u) =

  0

if |u(x) − u(y)| ≤ 1, for x − y ∈ supp(J),

 +∞

in other case,

as p → ∞, in the sense of Mosco. First, let us check that pepe

(11.16)

Epi(GJ∞ ) ⊂ s − lim inf Epi(GJp ). p→∞

J and λ ≥ 0 (as GJ∞ (u) = 0). To this end let (u, λ) ∈ Epi(GJ∞ ). We can assume that u ∈ K∞ Now take

def.up

(11.17)

vp = uχBR(p) (0)

and

λp = GJp (up ) + λ.

178


Then, as λ ≥ 0 we have (vp , λp ) ∈ Epi(GJp ). It is obvious that if R(p) → ∞ as p → ∞ we have vp → u in L2 (RN ), 1

and, if we choose R(p) = p 4N we get Z Z R(p)2N 1 J J (x − y) |vp (y) − vp (x)|p dy dx ≤ C Gp (vp ) = → 0, 2p RN RN p pepe

as p → ∞, we get (11.16). Finally, let us prove that e4T1.2

w − lim sup Epi(GJp ) ⊂ Epi(GJ∞ ).

(11.18)

p→∞

To this end, let us consider a sequence (upj , λpj ) ∈ Epi(GJpj ) (pj → ∞), that is, GJpj (upj ) ≤ λpj , with upj * u,

and

λpj → λ.

Therefore we obtain that 0 ≤ λ, since 0 ≤ GJpj (upj ) ≤ λpj → λ. On the other hand, we have that µZ Z ¶1/pj ¯ ¯pj ¯ ¯ J (x − y) upj (y) − upj (x) dy dx ≤ (Cpj )1/pj . RN

RN

Now, fix a bounded domain Ω ⊂ RN and q < pj . Then, by the above inequality, µZ Z ¶1/q ¯ ¯q ¯ ¯ J (x − y) upj (y) − upj (x) dy dx Ω

Ω

¶(pj −q)/pj q

µZ Z J (x − y) dy dx

≤ Ω

Ω

µZ

Z

× RN

µZ Z ≤

RN

¯ ¯p J (x − y) ¯upj (y) − upj (x)¯ j dy dx

J (x − y) dy dx Ω

¶1/pj

¶(pj −q)/pj q (Cpj )1/pj .

Ω

Hence, we can extract a subsequence (if necessary) and let pj → ∞ to obtain ¶1/q µZ Z ¶1/q µZ Z q . ≤ J (x − y) dy dx J (x − y) |u(y) − u(x)| dy dx Ω

Ω

Ω

Ω


179

Now, just taking q → ∞, we get |u(x) − u(y)| ≤ 1 As Ω is arbitrary we conclude that

a.e. (x, y) ∈ Ω × Ω, x − y ∈ supp(J). J u ∈ K∞ .


¤

0.32. Limit as ε → 0. Our next step is to rescale the kernel J appropriately and take the limit as the scaling parameter goes to zero. In the sequel we assume that supp(J) = B 1 (0). For given p > 1 and J we consider the rescaled kernels Z CJ,p ³ x ´ 1 −1 Jp,ε (x) := p+N J , where CJ,p := J(z)|zN |p dz ε ε 2 RN is a normalizing constant in order to obtain the p-Laplacian in the limit instead a multiple J of it. Associated to these kernels we have solutions up,ε to the nonlocal problems Pp p,ε (u0 ). AMRTp-Lap Let us also consider the solution to the local problem Pp (u0 ). Working as in [9] again, we can prove the following result.

vergencia.2

Theorem 150. Let p > N and assume J(x) ≥ J(y) if |x| ≤ |y|. Let T > 0, f ∈ J BV (0, T ; Lp (RN )), u0 ∈ Lp (RN ) and up,ε the unique solution of Pp p,ε (u0 ). Then, if vp is the unique solution of Pp (u0 ), lim sup kup,ε (t, ·) − vp (t, ·)kLp (RN ) = 0.

ε→0 t∈[0,T ]

AMRTp-Lap

BBM

We will use the following result from [9] which is a variant of [23, Theorem 4] (the first AMRTp-Lap statement is given in [9] for bounded domains Ω but it also holds for general open sets). prop:01.2

Proposition 151. Let Ω an open subset of RN . Let 1 ≤ p < +∞. Let ρ : RN → R be a nonnegative continuous radial function with compact support, non-identically zero, and ρn (x) := nN ρ(nx). Let {fn } be a sequence of functions in Lp (Ω) such that Z Z 1 |fn (y) − fn (x)|p ρn (y − x) dx dy ≤ M p . n Ω Ω (1) If {fn } is weakly convergent in Lp (Ω) to f , then f ∈ W 1,q (Ω) and moreover ¢ ¶ ¡ µ 1 x + f z − fn (x) 1 n n (ρ(z))1/p χΩ x + z * (ρ(z))1/p z · ∇f n 1/n weakly in Lp (Ω) × Lp (Ω). (2) If we further assume Ω is a smooth bounded domain in RN and ρ(x) ≥ ρ(y) if |x| ≤ |y| then {fn } is relatively compact in Lp (Ω), and consequently, there exists a subsequence {fnk } such that fnk → f in Lp (Ω) with f ∈ W 1,p (Ω). Using the above proposition we can take the limit as ε → 0 for a fixed p > N .

180


teo.3.intro.convergencia.2

Proof of Theorem 150. Recall that we have p > N and J(x) ≥ J(y) if |x| ≤ |y|, J T > 0, f ∈ L1 (0, T ; Lp (RN )), u0 ∈ Lp (RN ) and up,ε the unique solution of Pp p,ε (u0 ). We want to show that, if up is the unique solution of Np (u0 ), then font2

(11.19)

lim sup kup,ε (t, ·) − up (t, ·)kLp (RN ) = 0.

ε→0 t∈[0,T ]

font2

BCP

Cr

Since BpJ is m-accretive, to get (11.19) it is enough to see (see [20] or [49]) ¡ ¢−1 φ → (I + Bp )−1 φ in Lp (RN ) as ε → 0 I + BpJp,ε for any φ ∈ Cc (RN ).

ecu:01:b.2

¡ ¢−1 Let φ ∈ Cc (RN ) and uε := I + BpJp,ε φ. Then, µ ¶ Z Z Z CJ,p x−y uε v − p+N J |uε (y) − uε (x)|p−2 × ε ε N N N R R R Z (11.20) ×(uε (y) − uε (x)) dy v(x) dx =

φv RN

for every v ∈ L1 (RN ) ∩ L∞ (RN ). EEecu:01.2

9EEecu:01.2

Changing variables, we get (11.21) ¶ µ Z Z CJ,p x−y − p+N |uε (y) − uε (x)|p−2 (uε (y) − uε (x)) dy v(x) dx J ε ε RN RN ¯ ¯ Z Z ¯ uε (x + εz) − uε (x) ¯p−2 uε (x + εz) − uε (x) CJ,p ¯ = J(z) ¯¯ × ¯ 2 ε ε N N R

R

× ecu:01:b.2

v(x + εz) − v(x) dx dz. ε

So we can rewrite (11.20) as (11.22) Z Z φ(x)v(x) dx − RN

Z

Z

= RN

uε (x)v(x) dx RN

RN

¯ ¯ ¯ uε (x + εz) − uε (x) ¯p−2 uε (x + εz) − uε (x) CJ,p ¯ J(z) ¯¯ × ¯ 2 ε ε v(x + εz) − v(x) dx dz. × ε

We shall see that there exists a sequence εn → 0 such that uεn → u in Lp (RN ), u ∈ W 1,p (RN ) and u = (I + Bp )−1 φ, that is, Z Z Z p−2 uv + |∇u| ∇u · ∇v = φv for every v ∈ W 1,p (RN ) ∩ L∞ (RN ). RN

RN

RN


181

Since uε ¿ φ, there exists a sequence εn → 0 such that weakly in Lp (RN ),

uεn * u,

u ¿ φ. 99EEecu:01.2

ecu:02.2

Observe that kuεn kL∞ (RN ) , kukL∞ (RN ) ≤ kφkL∞ (RN ) . Taking ε = εn and v = uεn in (11.22) and applying Young’s inequality, we get ¯p µ ¶¯ Z Z 1 CJ,p x − y ¯¯ uεn (y) − uεn (x) ¯¯ J ¯ ¯ dx dy N εn εn RN RN 2 εn ¯ ¯ (11.23) Z Z Z ¯ uεn (x + εn z) − uεn (x) ¯p 1 CJ,p ¯ dx dz ≤ M := |φ(x)|2 dx. = J(z) ¯¯ ¯ εn 2 RN RN RN 2 prop:01.2

Therefore, by Proposition 151,

u ∈ W 1,p (RN ), uεn → u in Lploc (RN ) and µ ecua:03.22

(11.24)

¶1/p µ ¶1/p CJ,p uεn (x + εn z) − uεn (x) CJ,p J(z) * J(z) z · ∇u(x) 2 εn 2

weakly in Lp (RN ) × Lp (RN ). Let us prove now the tightness of {uεn }, which is to say, that no mass moves to infinity as p → +∞. For this, assume supp(φ) ⊂ BR (0) and fix S > 2R. Select a smooth function ϕ ∈ C ∞ (RN ) 99EEecu:01.2 such that 0 ≤ ϕ ≤ 1, ϕ ≡ 0 on BR (0), ϕ ≡ 1 on RN \ BS (0) and |∇ϕ| ≤ S2 . Taking in (11.22) ϕ|uεn |p−2 uεn , and having in mind that kuεn kL∞ (RN ) ≤ kφkL∞ (RN ) , we get Z |uεn |p (x)ϕ(x) dx RN Z Z CJ,p ≤ J(z)|uεn (x + εn z) − uεn (x)|p−1 |uεn (x + εn z)|p−1 p 2εn RN B1 (0) ×|ϕ(x + εn z) − ϕ(x)| dzdx ¯ ¯ ¯ uεn (x + εn z) − uεn (x) ¯p−1 ¯ ¯ dy dx J(z) ¯ ¯ ε n {|x|≤S+1} B1 (0) ¯ ¯ ¶ 10 µZ Z p ¯ uεn (x + εn z) − uεn (x) ¯p CJ,p kφkp−1 L∞ ¯ dy ≤ J(z) ¯¯ ¯ S εn {|x|≤S+1} B1 (0) 1 µZ ¶ Z p × J(z)dz dx CJ,p kφkp−1 L∞ ≤ S

= O(S

−1+ N p

),

Z

Z

{|x|≤S+1}

B1 (0)

182


ecu:02.2

the last equality being true by (11.23) and since µZ ¶ p1 Z N J(z)dz dx ≤ C(S + 1) p . {|x|≤S+1}

Consequently,

B1 (0)

Z

N

|uεn |p (x) dx = O(S −1+ p ) {|x|≥S}

uniformly in εn . Therefore, uεn → u in Lp (RN ). ecu:02.2

Moreover, from (11.23), we can also assume that ¯ ¯ ¯ uεn (x + εn z) − uεn (x) ¯p−2 uεn (x + εn z) − uεn (x) ¯ ¯ * χ(x, z) ¯ ¯ εn εn 0

ecu:0101.2

99EEecu:01.2

0

weakly in Lp (RN ) × Lp (RN ). Therefore, passing to the limit in (11.22) for ε = εn , we get Z Z Z Z CJ,p uv + J(z)χ(x, z) z · ∇v(x) dx dz = φv (11.25) RN RN RN 2 RN for every v smooth and by approximation for every v ∈ W 1,p (RN ) ∩ L∞ (RN ). AMRTp-Lap

ecua:02.2

From now on we follow closely the arguments in [9], but we include some details here for the sake of completeness. Let us see now that Z Z Z CJ,p (11.26) J(z)χ(x, z)z · ∇v(x) dx dz = |∇u|p−2 ∇u · ∇v. RN RN 2 RN 99EEecu:01.2ecu:0101.2

In fact, taking v = uεn in (11.22), by (11.25) we have ¯ ¯ Z Z Z Z ¯ uεn (x + εn z) − uεn (x) ¯p CJ,p ¯ ¯ J(z) ¯ ¯ dx dz = N φuεn − N uεn uεn εn RN Rn 2 R R Z Z Z Z Z = φu − uu − φ(u − uεn ) + 2u(u − uεn ) − (u − uεn )(u − uεn ) RN Rn RN RN RN Z Z Z Z CJ,p ≤ J(z)χ(x, z) z · ∇u(x) dx dz − φ(u − uεn ) + 2u(u − uεn ). RN RN 2 RN Rn N Consequently,

¯ ¯ ¯ uεn (x + εn z) − uεn (x) ¯p CJ,p ¯ ¯ dx dz lim sup J(z) ¯ ¯ 2 ε N N n n R Z RZ CJ,p ≤ J(z)χ(x, z) z · ∇u(x) dx dz. RN RN 2 Z

new:04.2

(11.27)

Z


183

DESIGU1.2

Now, by the monotonicity property (11.4), for every ρ smooth, ¶ µ Z Z CJ,p x−y |ρ(y) − ρ(x)|p−2 (ρ(y) − ρ(x)) dy (uεn (x) − ρ(x)) dx − p+N J εn ε N N n R R µ ¶ Z Z CJ,p x−y ≤ − p+N J |uεn (y) − uεn (x)|p−2 (uεn (y) − uεn (x)) dy (uεn (x) − ρ(x)) dx. εn εn RN RN EEecu:01.2

ecua:03.22

new:04.2

Using the change of variable (11.21) and taking limits, on account of (11.24) and (11.27), we obtain for every ρ smooth, Z Z CJ,p J(z)|z · ∇ρ|p−2 z · ∇ρ z · (∇u − ∇ρ) 2 N N R Z RZ CJ,p ≤ J(z)χ(x, z) z · (∇u(x) − ∇ρ(x)) dx dz, RN RN 2 and then, by approximation, for every ρ ∈ W 1,p (RN ). Taking now, ρ = u ± λv, λ > 0 and v ∈ W 1,p (RN ), and letting λ → 0, we get Z Z CJ,p J(z)χ(x, z)z · ∇v(x) dx dz RN RN 2 Z Z CJ,p = J(z) |z · ∇u(x)|p−2 (z · ∇u(x)) (z · ∇v(x)) dx dz. RN 2 Ω Consequently, Z Z Z CJ,p J(z)χ(x, z)z · ∇v(x) dx dz = CJ,p a(∇u) · ∇v RN RN 2 RN where

Z aj (ξ) = CJ,p RN

for every v ∈ W 1,p (RN ),

1 J(z) |z · ξ|p−2 z · ξ zj dz. 2

Hence, if we prove that E1AP.2

a(ξ) = |ξ|p−2 ξ,

(11.28) ecua:02.2

then (E1AP.2 11.26) is true and u = (I + Bp )−1 φ. So, to finish the proof we only need to show that (11.28) holds. Obviously, a is positively homogeneous of degree p − 1, that is, a(tξ) = tp−1 a(ξ)

for all ξ ∈ RN

and all t > 0.

E1AP.2

Therefore, in order to prove (11.28) it is enough to see that ai (ξ) = ξi

for all ξ ∈ RN , |ξ| = 1,

i = 1, . . . , N.

t Now, let Rξ,i be the rotation such that Rξ,i (ξ) = ei , where ei is the vector with components t (ei )i = 1, (ei )j = 0 for j 6= i, being Rξ,i the transpose of Rξ,i . Observe that t t t ξi = ξ · ei = Rξ,i (ξ) · Rξ,i (ei ) = ei · Rξ,i (ei ).

a1.ep.intro

184


−1 = On the other hand, since J is radial, CJ,p

a(ei ) = ei

1 2

R RN

J(z)|zi |p dz and

for every i.

Making the change of variables z = Rξ,i (y), since J is a radial function, we obtain Z 1 ai (ξ) = CJ,p J(z)|z · ξ|p−2 z · ξ z · ei dz 2 N ZR 1 t (ei ) dy = CJ,p J(y)|y · ei |p−2 y · ei y · Rξ,i 2 N R t t (ei ) = ξi , (ei ) = ei · Rξ,i = a(ei ) · Rξ,i and the proof finishes.

¤

0.33. The limit as ε → 0 in the sandpile model. Let us rescale the limit problem J P∞ (u0 ) considering the functionals   0 if u ∈ L2 (RN ), |u(x) − u(y)| ≤ ε, for |x − y| ≤ ε, ε G∞ (u) =  +∞ in other case, and the gradient flow associated to this functional,   f (t, ·) − ut (t, ·) ∈ ∂Gε∞ (u(t)), ε P∞ (u0 )  u(0, x) = u (x). 0

a.e t ∈]0, T [,

We have the following theorem. Theorem 152. Let T > 0, f ∈ L1 (0, T ; L2 (RN )), u0 ∈ L2 (RN ) ∩ W 1,∞ (RN ) such that ε k∇u0 k∞ ≤ 1 and consider u∞,ε the unique solution of P∞ (u0 ). Then, if v∞ is the unique solution of P∞ (u0 ), we have lim sup ku∞,ε (t, ·) − v∞ (t, ·)kL2 (RN ) = 0.

ε→0 t∈[0,T ]

AEW

EFG

Hence, we have approximated the sandpile model described in [10] and [56] by a nonlocal equation. In this nonlocal approximation a configuration of sand is stable when its height u verifies |u(x) − u(y)| ≤ ε when |x − y| ≤ ε. This is a sort of measure of how large is the size of irregularities of the sand; the sand can be completely irregular for sizes smaller than ε but it has to be arranged for sizes greater than ε. For ε > 0, we rescale the functional GJ∞ as follows   0 if |u(x) − u(y)| ≤ ε, for |x − y| ≤ ε, Gε∞ (u) =  +∞ in other case. In other words, Gε∞ = IKε , where Kε := {u ∈ L2 (RN ) : |u(x) − u(y)| ≤ ε, for |x − y| ≤ ε}.


185

Consider the gradient flow associated to the functional Gε∞   f (t, ·) − ut (t, ·) ∈ ∂IKε (u(t)), a.e. t ∈]0, T [, ε P∞ (u0 )  u(0, x) = u (x), in RN , 0 and the problem P∞ (u0 ) where

  f (t, ·) − u∞,t ∈ ∂IK0 (u∞ ),

a.e. t ∈]0, T [,

 u (0, x) = u (x), ∞ 0

in RN ,

© ª K0 := u ∈ L2 (RN ) ∩ W 1,∞ (RN ) : |∇u| ≤ 1 .

Observe that if u ∈ K0 , |∇u| ≤ 1. Hence, |u(x) − u(y)| ≤ |x − y|, from where it follows that u ∈ Kε , that is, K0 ⊂ Kε . With all these definitions and notations, we can proceed with the limit as ε → 0 for the sandpile model (p = ∞). convergencia1.ep.intro

Proof of Theorem 152. We have T > 0, f ∈ L1 (0, T ; L2 (RN )), u0 ∈ K0 and u∞,ε the ε unique solution of P∞ (u0 ). We have to show that if v∞ is the unique solution of P∞ (u0 ), then lim sup ku∞,ε (t, ·) − v∞ (t, ·)kL2 (RN ) = 0. ε→0 t∈[0,T ]

Since u0 ∈ K0 , u0 ∈ Kε for all ε > 0, and consequently there exists u∞,ε the unique ε solution of P∞ (u0 ). convergencia1.Mosco

By Theorem 146 to prove the result it is enough to show that IKε converges to IK0 in the sense of Mosco. It is easy to see that e0T1

(11.29)

Kε1 ⊂ Kε2 ,

if ε1 ≤ ε2 .

Since K0 ⊂ Kε for all ε > 0, we have K0 ⊂

\

Kε .

ε>0

On the other hand, if u∈

\

Kε ,

ε>0

we have

e1T1

|u(y) − u(x)| ≤ 1, a.e x, y ∈ RN , |y − x| from where it follows that u ∈ K0 . Therefore, we have \ (11.30) K0 = Kε . ε>0

186


Note that e2T1

(11.31)

e1T1

Epi(IK0 ) = K0 × [0, ∞[, e2T1

Epi(IKε ) = Kε × [0, ∞[ ∀ ε > 0.

By (11.30) and (11.31), we have e3T1

(11.32)

Epi(IK0 ) ⊂ s − lim inf Epi(IKε ). ε→0

On the other hand, given (u, λ) ∈ w − lim supε→0 Epi(IKε ) there exists (uεk , λk ) ∈ Kεk × [0, ∞[, such that εk → 0 and uεk * u in L2 (RN ),

e0T1

λk → λ in R.

By (11.29), given ε > 0, there exists k0 , such that u ∈ Kε for all k ≥ k0 . Then, since Kε is e1T1 εk a closed convex set, we get u ∈ Kε , and, by (11.30), we obtain that u ∈ K0 . Consequently, e4T1

psing.intro

(11.33)

w − lim sup Epi(IKε ) ⊂ Epi(IK0 ). e3T1

e4T1

n→∞

e1Mos

Finally, by (11.32) and (11.33), and having in mind (11.1), we obtain that IKε converges to IK0 in the sense of Mosco. ¤ 0.34. Collapse of the initial condition. EFG

In [56] the authors studied the collapsing of the initial condition phenomena for the local problem Pp (u0 ) when the initial condition u0 satisfies k∇u0 k∞ > 1. They find that the limit of the solutions vp (x, t) to Pp (u0 ) is independent of time but does not coincide with u0 . They also describe the small layer in which the solution rapidly changes from being u0 at t = 0 to something close to the final stationary limit for t > 0. Now, our task is to perform a similar analysis for the nonlocal problem. To this end let us take ε = 1 and f = 0 and look for the limit as p → ∞ of the solutions to the nonlocal problem up when the initial condition u0 does not verify that |u0 (x) − u0 (y)| ≤ 1 for x − y ∈ supp(J). We get that the nonlinear nature of the problem creates an initial short-time layer in which the solution changes very rapidly. We describe this layer by means of a limit evolution problem. We have the following result. Theorem 153. Let up be the solution to PpJ (u0 ) with initial condition u0 ∈ L2 (RN ) such that 1 1. They prove that for each time t > 0 vp (t, ·) → v∞ (·),

uniformly as p → +∞,

where v∞ is independent of time and satisfies ess supRN |∇v∞ | ≤ 1. Moreover, v∞ (x) = v(1, x), v solving the nonautonomous evolution equation  v  − vt ∈ ∂IK0 (v), t ∈]τ, ∞[ t  v(τ, x) = τ u0 (x), where τ = L−1 . They interpreted this as a crude model for the collapse of a sandpile from an initially unstable configuration. The proof of this result BEG is based in a scaling argument, which was extended by Bénilan, Evans and Gariepy in [21], to cover general nonlinear evolution equations governed by homogeneous accretive operators. Here, using this general result, we prove similar results for our nonlocal model. We look for the limit as p → ∞ of the solutions to the nonlocal problem PpJ (u0 ) when the initial datum u0 satisfies 1 2, we consider in the Banach space X = L2 (RN ) the operators ∂GJp . Then, ∂GJp are m-accretive operators in L2 (Rn ) and also positively homogeneous of degree p − 1.

ompactness1

188


Moreover, the solution up to the nonlocal problem PpJ (u0 ) coincides with the strong solution of the abstract Cauchy problem   −ut (x, t) ∈ ∂GJp (u(t)), a.e t ∈]0, T [,  u(0, x) = u (x), 0 Let

x ∈ RN .

© ª C := u ∈ L2 (RN ) : ∃(up , vp ) ∈ ∂GJp with up → u, vp → 0 as p → ∞ .

It is easy to see that © ª J C = K∞ = u ∈ L2 (RN ) : |u(x) − u(y)| ≤ 1, for x − y ∈ supp(J) . Then, X0 :=

[

L2 (RN )

λC

= L2 (RN ).

λ>0

Lemma 154. For f ∈ L2 (RN ) and p > N , let up := (I + ∂GJp )−1 f . Then, the set of functions {up : p > N } is precompact in L2 (RN ). Proof. First assume that f is bounded and the support of f lies in the ball BR (0). Since the operator ∂GJp is completely accretive, observe that L2 (RN )

∂GJp = BpJ ∩ (L2 (RN ) × L2 (RN ))

,

we have the estimates kup kL∞ ≤ kf kL∞ ,

kup kL2 ≤ kf kL2

and kup (·) − up (· + h)kL2 ≤ kf (·) − f (· + h)kL2 for each h ∈ RN . Consequently, {up : p > N } is precompact in L2 (K) for each compact set K ⊂ RN . We must show that {up : p > N } is tight. For this, fix S > 2R and select a smooth function ϕ ∈ C ∞ (RN ) such that 0 ≤ ϕ ≤ 1, ϕ ≡ 0 on BR (0), ϕ ≡ 1 on RN \ BS (0) and |∇ϕ| ≤ S2 . We have Z up (x) = J(x − y)|up (y) − up (x)|p−2 (up (y) − up (x)) dy + f (x). RN


189

Then, multiplying by ϕup and integrating, we get Z u2p (x)ϕ(x) dx RN Z Z = J(x − y)|up (y) − up (x)|p−2 (up (y) − up (x))up (x)ϕ(x) dydx N N R ZR Z 1 J(x − y)|up (y) − up (x)|p−2 (up (y) − up (x))(up (y)ϕ(y) − up (x)ϕ(x)) dydx =− 2 RN RN Z Z 1 J(x − y)|up (y) − up (x)|p−2 (up (y) − up (x))up (y)(ϕ(y) − ϕ(x)) dydx. ≤− 2 RN RN Now, since |∇ϕ| ≤ S2 , by Hölder’s inequality we obtain ¯ Z Z ¯ ¯ ¯1 p−2 ¯ ¯ ¯ 2 N N J(x − y)|up (y) − up (x)| (up (y) − up (x))up (y)(ϕ(y) − ϕ(x)) dydx¯ R R µZ ¶ Z kf kL∞ p−1 J(x − y)|up (y) − up (x)| dy dx ≤ S {|x|≤S+1} B1 (x) ¶ 10 µZ Z p kf kL∞ J(x − y)|up (y) − up (x)|p dy ≤ S {|x|≤S+1} B1 (x) ¶ p1 µZ Z J(x − y)dy × dx {|x|≤S+1}

≤ M (S + 1)

N p

−1

= O(S

B1 (x) −1+ N p

the last inequality being true since Hence, Z

), RR

J(x − y)|up (y) − up (x)|p is bounded uniformly in p. N

{|x|≥S}

u2p (x) dx = O(S −1+ p )

uniformly in p > N . This proves tightness and we have established compactness in L2 (RN ) provided f is bounded and has compact support. The general case follows, since such functions are dense in L2 (RN ). ¤ teo.collapsing.intro

Proof of Theorem 153. By the above Lemma, given f ∈ L2 (RN ) if up := (I + ∂GJp )−1 f , there exists a sequence pj → +∞, such that upj → v in L2 (RN ) as j → ∞. In convergencia.p.intro J , the proof of Theorem 145 we have established that the functionals GJp converge to IK∞ convergencia1.Mosco −1 J ) as p → ∞, in the sense of Mosco. Then, by Theorem 146, we have v = (I + IK∞ f. Therefore, the limit P f := lim (I + ∂GJp )−1 f p→∞

2

N

2

J . Moreover, exists in L (R ), for all f ∈ X0 = L (RN ), and P f = f if f ∈ C = K∞ J P −1 − I = ∂IK∞

190


and u = P f is the unique solution of J u 3 f. u + ∂IK∞

BEG

teo.collapsing.intro

Therefore, as consequence of the main result of [21], we have obtained Theorem 153.

¤

0.35. Explicit solutions. In this chapter we show some explicit examples of solutions to   f (x, t) − ut (x, t) ∈ ∂Gε∞ (u(t)), a.e. t ∈]0, T [, ε P∞ (u0 )  u(0, x) = u (x), in RN , 0 where Gε∞ (u)

=

  0

if u ∈ L2 (RN ), |u(x) − u(y)| ≤ ε, for |x − y| ≤ ε,

 +∞

in other case.

ε In order to verify that a function u(x, t) is a solution to P∞ (u0 ) we need to check that

ef.solution

(11.34)

Gε∞ (v) ≥ Gε∞ (u) + hf − ut , v − ui,

for all v ∈ L2 (RN ).

def.solution

To this end we can assume that v ∈ Kε (otherwise Gε∞ (v) = +∞ and then (11.34) becomes trivial). Therefore,

.solution.2

(11.35)

u(t, ·) ∈ Kε

def.solution

.solution.3

and (11.34) can be rewritten as Z (11.36) 0 ≥ (f (x, t) − ut (x, t))(v(x) − u(x, t)) dx R

for every v ∈ Kε . Example 1. Let us consider, in one space dimension, as source an approximation of a delta function 1 f (x, t) = fη (x, t) = χ[− η2 , η2 ] (x), 0 < η ≤ 2ε, η and as initial datum u0 (x) = 0.

ec.u.1

Now, let us find the solution by looking at its evolution between some critical times. ε (u0 ) is given by First, for small times, the solution to P∞ t (11.37) u(x, t) = χ[− η2 , η2 ] (x), η for t ∈ [0, ηε). Remark that t1 = ηε is the first time when u(x,def.solution.3 t) = ε and hence it is immediate that u(t, ·) ∈ Kε . Moreover, as ut (x, t) = f (x, t) then (11.36) holds.


191

For times greater than t1 the support of the solution is greater than the support of f . Indeed the solution can not be larger than ε in [− η2 , η2 ] without being larger than zero in the adjacent intervals of size ε, [ η2 , η2 + ε] and [− η2 − ε, − η2 ]. We have

ec.u.2

 ε + k1 (t − t1 )     k1 (t − t1 ) u(x, t) =     0

(11.38)

for x ∈ [− η2 , η2 ], for x ∈ [− η2 − ε, η2 + ε] \ [− η2 , η2 ], for x ∈ / [− η2 − ε, η2 + ε],

for times t such that t ∈ [t1 , t2 ) where k1 =

1 2ε + η

and

t2 = t1 +

ε = 2ε2 + 2εη. k1

Note that t2 is the first time when u(x, t) = 2ε for x ∈ [− η2 , η2 ]. Again it is immediate to see that u(t, ·) ∈ Kε , since fordef.solution.3 |x − y| < ε the maximum of the difference u(x, t) − u(y, t) is exactly ε. Now let us check (11.36). ec.u.2

julio

Using the explicit formula for u(x, t) given in (11.38), we obtain (11.39) ¶ Z Z η µ 2 1 (f (x, t) − ut (x, t))(v(x) − u(x, t)) dx = − ut (x, t) (v(x) − u(x, t)) dx η − η2 R Z −η Z η +ε 2 2 (−ut (x, t))(v(x) − u(x, t)) dx + (−ut (x, t))(v(x) − u(x, t)) dx + Z =

η 2

η 2

− η2

Z

+ µ =

µ

1 − k1 (v(x) − (ε + k1 (t − t1 ))) dx + η

− η2

− η2 −ε

µ

−η Z −

− η2 −ε

¶

η +ε 2 η 2

(−k1 )(v(x) − (k1 (t − t1 ))) dx

(−k1 )(v(x) − (k1 (t − t1 ))) dx

1 − k1 η η +ε 2

η 2

Z

¶

¶

µ

+ 2εk1 k1 (t − t1 ) − εη

k1 v(x) dx −

Z

− η2 − η2 −ε

¶ Z η µ ¶ 2 1 1 − k1 + − k1 v(x) dx η η − η2

k1 v(x) dx.

From our choice of k1 we get µ −η

¶ 1 − k1 + 2εk1 = 0 η

192


and, since v ∈ Kε , we have

Z (f (x, t) − ut (x, t))(v(x) − u(x, t)) dx R

FuenJul1

(11.40)

2εk1 = −2ε k1 + η

Z

η 2

2

− η2

Z v(x) dx − k1

η +ε 2 η 2

Z v(x) dx − k1

− η2 − η2 −ε

v(x) dx ≤ 0.

In fact, without loss of generality we can suppose that Z η 2 v(x) dx = 0. − η2

Then ec:2ep1

Z

Z

η/2

(11.41)

0

(−v) = a, 0

(−v) = −a. −η/2

Consequently, ec:2ep2

2 −v ≤ a + ε in [0, ε]. η

(11.42) ec:2ep2

ec:2ep1

Indeed, if (11.42) does not hold, then −v > η2 a in [0, ε] which contradicts (11.41). ec:2ep1

ec:2ep3day

Now, by (11.41), since v ∈ Kε , Z ε+η/2 Z η/2 (−v(x))dx = (−v(y + ε))dy ε 0 Z η/2 Z η/2 (11.43) = (−v(y + ε) + v(y))dy + (−v(y))dy 0

ec:2ep2

ec:2ep3

ec:2ep3day

0

Therefore, by (11.42) and (11.43), Z ε+η/2 Z ε Z ε+η/2 (−v) = (−v) + (−v) ε η/2 η/2 µ ¶³ (11.44) 2 η´ η 2 ≤ a+ε ε− + ε + a = aε + ε2 . η 2 2 η Similarly,

ec:2ep5

η ≤ ε + a. 2

Z

(11.45) ec:2ep3

−η/2

2 (−v) ≤ − aε + ε2 . η −ε−η/2

ec:2ep5

Consequently, by (11.44) and (11.45), Z η +ε Z 2 (−v) + η 2

− η2 − η2 −ε

(−v) ≤ 2ε2 .


.u.gener.ww

193

Now, it is easy to generalize and verify the following general formula that describes the solution for every t ≥ 0. For any given integer l ≥ 0 we have (11.46)  lε + kl (t − tl ), x ∈ [− η2 , η2 ],       (l − 1)ε + k (t − t ), x ∈ [− η − ε, η + ε] \ [− η , η ],   l l 2 2 2 2    ... u(x, t) =      kl (t − tl ), x ∈ [− η2 − lε, η2 + lε] \ [− η2 − (l − 1)ε, η2 + (l − 1)ε],       0, x∈ / [− η2 − lε, η2 + lε], for t ∈ [tl , tl+1 ), where kl =

1 2lε + η

and tl+1 = tl +

ε , kl

t0 = 0.

ec.u.gener.ww

From formula (11.46) we get, taking the limit as η → 0, that the expected solution to (11.34) with f = δ0 is given by, for any given integer l ≥ 1,  (l − 1)ε + kl (t − tl ), x ∈ [−ε, ε],        (l − 2)ε + kl (t − tl ), x ∈ [−2ε, 2ε] \ [−ε, ε],     ... (11.47) u(x, t) =       kl (t − tl ), x ∈ [−lε, lε] ∪ [−(l − 1)ε, (l − 1)ε],      0, x∈ / [−lε, lε], NOnLOpepsi.sec.5

ec.u.gener

for t ∈ [tl , tl+1 ) where kl =

1 , 2lε

tl+1 = tl +

ε , kl

t1 = 0.

Remark that, since the space of functions Kε is not contained into C(R), the formulation def.solution.3 ec.u.gener the function u(x, t) described by (11.47) is (11.36) with f = δ0 does not make sense. HenceNOnLOpepsi.sec.5 to be understood as a generalized solution to (11.34) (it is obtained as a limit of solutions to approximating problems). Note that the function u(tl , x) is a “regular and symmetric pyramid” composed by squares of side ε.

194


Recovering the sandpile model as ε → 0. Now, to recover the sandpile model, let us fix lε = L, and take the limit as ε → 0 in the previous example. We get that u(x, t) → v(x, t), where v(x, t) = (L − |x|)+ ,

for t = L2 ,

that is exactly the evolution given by the sandpile model with initial datum u0 = 0 and a AEW point source δ0 , see [10].

convergencia1.ep

Therefore, this concrete example illustrates the general convergence result Theorem 152. ec.u.gener.ww

Example 2. The explicit formula (11.46) can be easily generalized to the case in where the source depends on t in the form f (x, t) = ϕ(t)χ[− η2 , η2 ] (x), with ϕ a nonnegative integrable function and 0 < η ≤ ε. We arrive to the following formulas, setting Z t ϕ(s)ds, g(t) = 0

for any given integer l ≥ 0,   lε + kˆl (g(t) − g(tl )) ,        (l − 1)ε + kˆl (g(t) − g(tl )) ,     u(x, t) = . . .       kˆl (g(t) − g(tl )) ,       0,

x ∈ [− η2 , η2 ], x ∈ [− η2 − ε, η2 + ε] \ [− η2 , η2 ],

x ∈ [− η2 − lε, η2 + lε] \ [− η2 − (l − 1)ε, η2 + (l − 1)ε] x∈ / [− η2 − lε, η2 + lε],

for t ∈ [tl , tl+1 ), where kˆl =

η η + 2lε

and g(tl+1 ) − g(tl ) =

ε , kˆl

t0 = 0.

Observe that tl is the first time at which the solution reaches level lε. We can also consider ϕ changing sign. In this case the solution increases if ϕ(t) is positive in every interval of size ε (around the support of the source [− η2 , η2 ]) for which u(x) − u(y) = iε with |x − y| = iε for some x ∈ [− η2 , η2 ] (here i is any integer). While if ϕ(t) is negative the solution decreases in every interval of size ε for which u(x) − u(y) = −iε with |x − y| = iε for some x ∈ [− η2 , η2 ].


195

ec.u.2

nofun

def.solution.3

Example 3. Observe that if η > 2ε, then u(x, t) given in (11.38) does not satisfy (11.36) for a test function v ∈ Kε whose values in [− η2 − ε, η2 + ε] are  ε  −β + 2ε for x ∈ [− η2 + ε, η2 − ε],   2    ε −β + ε for x ∈ [− η2 , η2 ] \ [− η2 + ε, η2 − ε], v(x) = 2       −β ε for x ∈ [− η2 − ε, η2 + ε] \ [− η2 , η2 ], 2 for β = 4(1 − ε/η) which is greater that 2. At this point one can ask what happens in the previous situation when η > 2ε. In this case the solution begins to grow as before with constant speed in the support of f but after the first time when it reaches level ε the situation changes. Consider, for example, that the source is given by 1 f (x, t) = χ[−2ε,2ε] (x). ε In this case the solution to our nonlocal problem with u0 (x) = 0, u(x, t), can be described as follows. Firstly we have t u(x, t) = χ[−2ε,2ε] (x), ε for t ∈ [0, ε2 ). Remark that t1 = ε2 is the first time when u(x,def.solution.3 t) = ε and hence it is immediate that u(t, ·) ∈ Kε . Moreover, as ut (x, t) = f (x, t) then (11.36) holds. For times greater that t1 we have  1  ε + (t − t1 )    ε     ε + k1 (t − t1 ) u(x, t) =    k1 (t − t1 )      0

for x ∈ [−ε, ε], for x ∈ [−2ε, −ε] ∪ [ε, 2ε], for x ∈ [−3ε, −2ε] ∪ [2ε, 3ε], for x ∈ / [−3ε, 3ε],

for t ∈ [t1 , t2 ), where

1 and t2 = ε2 + 2ε2 = 3ε2 . 2ε def.solution.3 With this expression of u(x, t) it is easy to see that it verifies (11.36). k1 =

ec.u.gener.ww

For times greater than t2 an expression similar to (11.46) holds. We leave the details to the reader.

196


Example 4. For two or more dimensions we can get similar formulas. Given a bounded domain Ω0 ⊂ RN let us define inductively © ª Ω1 = x ∈ RN : ∃y ∈ Ω0 , with |x − y| < ε and

© ª Ωj = x ∈ RN : ∃y ∈ Ωj−1 , with |x − y| < ε .

In the sequel, for simplicity, we consider the two dimensional case N = 2. Let us take as source f (x, t) = χΩ0 (x), Ω0 = B(0, ε/2), and, as initial datum, u0 (x) = 0.

.u.gener.RN

NOnLOpepsi.sec.5

In this case, for any integer l ≥ 0, the solution to (11.34) is given by   lε + kˆl (t − tl ), x ∈ Ω0 ,        (l − 1)ε + kˆl (t − tl ), x ∈ Ω1 \ Ω0 ,     (11.48) u(x, t) = ...     S   kˆl (t − tl ), x ∈ Ωl \ l−1  j=1 Ωj ,      0, x∈ / Ωl , for t ∈ [tl , tl+1 ), where

|Ω0 | ε kˆl = , tl+1 = tl + , t0 = 0. |Ωl | kˆl Note that the solution grows in strips of width ε around the set Ω0 where the source is localized. As in the previous examples, the result is evident for t ∈ [0, t1 ). Let us see it for t ∈ [t1 ,def.solution.3 t2 ), a similar argument works for later times. It is clear that u(t, ·) ∈ Kε , let us check (11.36). Working as in Example 1, we must show that Z Z ˆ ˆ (1 − k1 ) v − k1 v ≤ (1 − kˆ1 )ε|Ω0 | ∀v ∈ Kε , Ω0

mazon.2

Ω1 \Ω0

where Ω1 = B(0, 3ε/2). Since kˆ1 = |Ω0 |/|Ω1 |, the last inequality is equivalent to ¯ ¯ Z Z ¯ 1 ¯ 1 ¯ ¯ ≤ ε ∀v ∈ Kε . (11.49) v − v ¯ |Ω0 | |Ω1 \ Ω0 | Ω1 \Ω0 ¯ Ω0 mazon.2

By density, it is enough to prove (11.49) for any v ∈ Kε continuous.


197

Let us now divide Ω0 = {r(cos θ, sin θ) : 0 ≤ θ ≤ 2π, 0 ≤ r < 2ε } and Ω1 \ Ω0 = {r(cos θ, sin θ) : 0 ≤ θ ≤ 2π, ε ≤ r < 32 ε} as follows. Consider the partitions 0 = θ0 < θ1 < .... < θN = 2π, with θi − θi−1 = 2π/N , N ∈ N, 0 = r0 < r1 < .... < rN = ε/2 and ε/2 = r˜0 < r˜1 < .... < rÑ = 3ε/2, such that the measure of Bij = {r(cos θ, sin θ) : θi−1 < θ < θi rj−1 < r < rj } is constant, that is, |Bij | = |Ω0 |/N 2 , and the measure of Aij = {r(cos θ, sin θ) : θi−1 < θ < θi r˜j−1 < r < r˜j } is also constant, that is, |Aij | = |Ω1 \ Ω0 |/N 2 . In this way we have partitioned Ω0 and Ω1 \ Ω0 as a disjoint family of N 2 sets such that ¯ ¯ ¯ ¯ N N ¯ ¯ ¯ ¯ [ [ ¯ ¯ ¯ ¯ Ω \ B = 0, (Ω \ Ω ) \ A ¯ 0 ¯ 1 ij ¯ 0 ij ¯ = 0. ¯ ¯ ¯ ¯ i,j=1

i,j=1

By construction, if we take xij = rj (cos θi−1 , sin θi−1 ) ∈ Bij ,

x˜ij = r˜j−1 (cos θi−1 , sin θi−1 ) ∈ Aij ,

then |xij − x˜ij | ≤ ε for all i, j = 1, . . . N . Given a continuous function v ∈ Kε , by uniform continuity of v, for δ > 0, there exists ρ > 0 such that δ |v(x) − v(y)| ≤ if |x − y| ≤ ρ. 2 Hence, if we take N big enough such that diameter(Bij ) ≤ ρ and diameter(Aij ) ≤ ρ, we have ¯ ¯Z N ¯ δ|Ω | ¯ X ¯ ¯ 0 v(x) − v(xij )|Bij |¯ ≤ ¯ ¯ ¯ Ω0 2 i,j=1 and

¯ ¯Z N ¯ δ|Ω \ Ω | ¯ X ¯ ¯ 1 0 v(x) − v(˜ xij )|Aij |¯ ≤ . ¯ ¯ ¯ Ω1 \Ω0 2 i,j=1

Since v ∈ Kε and |xij − x˜ij | ≤ ε, |v(xij ) − v(˜ xij )| ≤ ε. Consequently,

198


¯ ¯ Z Z ¯ 1 ¯ 1 ¯ v− v ¯¯ ¯ |Ω0 | |Ω1 \ Ω0 | Ω1 \Ω0 Ω0 ¯ ¯ N N ¯ 1 X ¯ X 1 ¯ ¯ ≤¯ v(xij )|Bij | − v(˜ xij )|Aij |¯ + δ ¯ |Ω0 | ¯ |Ω1 \ Ω0 | i,j=1 i,j=1 ¯ ¯ N N ¯ 1 X ¯ 1 X ¯ ¯ =¯ 2 v(xij ) − 2 v(˜ xij )¯ + δ ¯N ¯ N i,j=1

≤ ε + δ.

i,j=1

mazon.2

Therefore, since δ > 0 is arbitrary, (11.49) is obtained. ec.u.gener.RN

Again the explicit formula (11.48) can be easily generalized to the case where the source depends on t in the form f (x, t) = ϕ(t)χΩ0 (x). An estimate of the support of ut . Taking a source f ≥ 0 supported in a set A, let us see where the material is added (places where ut is positive). Let us compute a set that we will call Ω∗ (t) as follows. Let Ω0 (t) = A, and define inductively © ª Ω1 (t) = x ∈ RN \ Ω0 (t) : ∃y ∈ Ω0 (t) with |x − y| < ε and u(y, t) − u(x, t) = ε and

© ª Ωj (t) = x ∈ RN \ Ωj−1 (t) : ∃y ∈ Ωj−1 (t) with |x − y| < ε and u(y, t) − u(x, t) = ε .

With these sets Ωi (t) (observe that there exists a finite number of such sets, since u(x, t) is bounded) let [ Ω∗ (t) = Ωi (t). i

We have that ut (x, t) = 0,

for x ∈ / Ω∗ (t).

def.solution.3

Indeed, this can be easily deduced using an appropriate test function v in (11.36). Just take v(x) = u(x, t) but for a small neighborhood near x ∈ / Ω∗ (t). Example 5. Finally, note that an analogous description like in the above examples can be made for an initial condition that is of the form K X u0 (x) = ai χ[iε,(i+1)ε] (x), i=−K

with |ai − ai±1 | ≤ ε,

a−K = aK = 0,


199

(this last condition is needed just to imply that u0 ∈ Kε ) together with the sum of a finite number of delta functions placed at points xl = lε (or a finite sum of functions of time times the characteristic functions of some intervals of the form [lε, (l + 1)ε]) as the source term. For example, let us consider a source placed in just one interval, f (x, t) = χ[0,ε] (x). Initially, u(0, x) = lε for x ∈ [0, ε]. Let us take w1 (x) the regular and symmetric pyramid centered at [0, ε] of height (l + 1)ε (and base of length (2l − 1)ε). With this pyramid and the initial condition let us consider the set Λ1 = {j : w1 (x) > u(0, x) for x ∈ (jε, (j + 1)ε)} . This set contains the indexes of the intervals in which the sand is being added in the first stage. During this first stage, u(x, t) is given by X t χ[jε,(j+1)ε] (x), u(x, t) = u(0, x) + Card(Λ1 ) j∈Λ 1

for t ∈ [0, t1 ], where t1 = Card(Λ1 )ε is the first time at which u is of size (l + 1)ε in the interval [0, ε]. From now on the evolution follows the same scheme. In fact, X t − ti χ[jε,(j+1)ε] (x), u(x, t) = u(ti , x) + Card(Λi ) j∈Λ i

for t ∈ [ti , ti+1 ], ti+1 − ti = Card(Λi )ε. Where, from the pyramid wi of height (l + i)ε, we obtain Λi = {j : wi (x) > u(ti , x) for x ∈ (jε, (j + 1)ε)} . Remark that eventually the pyramid wk is bigger than the initial condition, from this time on the evolution is the same as described for u0 = 0 in the first example. In case we have two sources, the pyramids wi , w˜i corresponding to the two sources eventually intersect. In the interval where the intersection takes place, ut is given by the greater of the two possible speeds (that correspond to the different sources). If both possible speeds are the same this interval has to be computed as corresponding to both sources simultaneously. Recovering the sandpile model. Note that any initial condition w0 with |∇w0 | ≤ 1 can an explicit be approximated by an u0 like the one described above. Hence we can obtain AEW solution of the nonlocal model that approximates the solutions constructed in [10]. Compact support of the solutions. Note also that when the source f and the initial condition u0 are compactly supported and bounded then also the solution is compactly supported and bounded for all positive times. This property has to be contrasted with the fact that solutions to the nonlocal p−laplacian PpJ (u0 ) are not compactly supported even if u0 is.

trans.intro

200


0.36. A mass transport interpretation. We can also give an interpretation ofEFG the limit problem P (u ) in terms of Monge-Kantorovich mass transport theory as in [56], Feldman Villani∞ 0 [58] (see [84] for a general introduction to mass transportation problems). To this end let us consider the distance   0 if x = y, d(x, y) =  [|x − y|] + 1 if x 6= y. Here [·] means the entire part of the number. Note that this function d measures distances with jumps of length one. Then, given two measures (that for simplicity we will take absolutely continuous with respect to the Lebesgue measure in RN ) f+ , f− in RN , and supposing the overall condition of mass balance Z Z + f − dy, f dx = RN

RN

the Monge’s problem associated to the distance d is given by: minimize Z d(x, s(x)) f+ (x)dx among the set of maps s that transport f+ into f− , which means Z Z + h(s(x))f (x) dx = h(y)f − (y) dy RN

RN

for each continuous function h : RN → Evans R. The dual formulation of this minimization problem, introduced by Kantorovich (see [55]), is given by Z max u(x)(f+ (x) − f− (x))dx u∈K∞

RN

where the set K∞ is given by K∞ := {u ∈ L2 (RN ) : |u(x) − u(y)| ≤ 1, for |x − y| ≤ 1}. We are assuming that supp(J) = B 1 (0) (in other case we have to redefine the distance d accordingly). With these definitions and notations we have the following result. J (u0 ) is a solution to the Theorem 155. The solution u∞ (t, ·) of the limit problem P∞ dual problem Z u(x)(f+ (x) − f− (x))dx max u∈K∞

RN

when the involved measures are the source term f+ = f (x, t) and the time derivative of the solution f− = ut (x, t).


201

teo.trans.intro

Proof of Theorem 155. Let d(x, y) =

  0

if x = y,

 [|x − y|] + 1

if x 6= y.

Here [·] means the entire part of the number. Note that this function d measures distances with jumps of length one. Then, given two measures (that for simplicity we will take absolutely continuous with respect to the Lebesgue measure in RN ) f+ , f− in RN , and supposing the overall condition of mass balance Z Z f + dx = RN

minimiz.M-K

maximiz

ion.3.monge

f − dy, RN

the Monge’s problem associated to the distance d is given by: minimize Z (11.50) d(x, s(x)) f+ (x)dx among maps s that transport f+ into f− . The dual formulation of this problem is given by Z (11.51) max u(x)(f+ (x) − f− (x))dx u∈K∞

RN

where, as before, K∞ is given by © ª K∞ := u ∈ L2 (RN ) : |u(x) − u(y)| ≤ 1, for |x − y| ≤ 1 . We are assuming that supp(J) = B 1 (0) (in other case we may redefine the distance d accordingly). Then it is easy to obtain maximiz that the solution u∞ (t, ·) of the limit problem GJ∞ (u0 ) is a solution to the dual problem (11.51) when the involved measures are the source f (x, t) and the time derivative of the solution u∞,t (x, t). In fact, we have GJ∞ (v) ≥ GJ∞ (u∞ ) + hf − u∞,t , v − u∞ i,

for all v ∈ L2 (RN ).

That is equivalent to u∞ (t, ·) ∈ K∞ and

Z

(11.52)

0≥

(f (x, t) − u∞,t (x, t))(v(x) − u∞ (x, t)) dx RN

def.solution.3.monge

for every v ∈ K∞ . Now, we just observe that (11.52) is Z Z (f (x, t) − u∞,t (x, t))u∞ (x, t) dx ≥ (f (x, t) − u∞,t (x, t))v(x) dx. RN

RN

Therefore, we have that u∞ (t, ·) is a solution to the dual mass transport problem.

202


Consequently, we conclude that the mass of sand added by the source f (t, ·) is transported (via u(t, ·) as the transport potential) to u∞,t (t, ·) at each time t. ¤ This mass transport interpretation of the problem can be clearly observed looking at the previous concrete examples. 0.37. Neumann boundary conditions. Finally, let us observe that analogous results are also valid when we consider the Neumann problem in a bounded convex domain Ω, that is, when all the involved integrals are taken in Ω. AMRTp-Lap

Let Ω be a convex domain in RN . As we have mentioned, in [9] we have studied the evolution problem Z   up,t (x, t) = J(x − y)|up (y, t) − up (x, t)|p−2 (up (y, t) − up (x, t))dy + f (x, t), J,Ω Pp (u0 ) Ω  up (0, x) = u0 (x), in Ω. The associated functional being GJ,Ω p (u)

1 = 2p

Z Z J(x − y)|u(y) − u(x)|p dy dx. Ω

Ω

This is the nonlocal analogous to the Neumann problem for the p−Laplacian since in this AMRTp-Lap evolution problem, we have imposed a zero flux condition across the boundary of Ω, see [9]. Also, let us consider the rescaled problems with Jε , that we call PpJε ,Ω (u0 ), and the corresponding limit problems   f (t, ·) − ut (t, ·) ∈ ∂GJ,Ω a.e. t ∈]0, T [, ∞ (u(t)), ε,Ω P∞ (u0 )  u(0, x) = u (x), in Ω. 0 With associated functionals   0 Gε,Ω (u) = ∞  +∞

if |u(x) − u(y)| ≤ ε, for |x − y| ≤ ε; x, y ∈ Ω, in other case.

The limit problem of the local p−Laplacians being  Ω  f (t) − v∞,t ∈ ∂F∞ (v∞ (t)), Ω P∞ (u0 )  v (0, x) = g(x), ∞

a.e. t ∈]0, T [, in Ω,

Ω is defined in L2 (Ω) by where the functional F∞   0 if |∇v| ≤ 1, Ω F∞ (v) =  +∞ in other case.

ect.Newmann


203

In these limit problems we assume that the material is confined in a domain Ω, thus Feldman we are looking at models for sandpiles inside a container, see also [58]. Working as in the previous sections we can prove that Theorem 156. Let Ω be a convex domain in RN . (1) Let T > 0, u0 ∈ L2 (Ω) such that |u0 (x) − u0 (y)| ≤ 1, for x − y ∈ Ω ∩ supp(J) and J,Ω up the unique solution of PpJ,Ω (u0 ). Then, if u∞ is the unique solution to P∞ (u0 ), lim sup kup (t, ·) − u∞ (t, ·)kL2 (Ω) = 0.

p→∞ t∈[0,T ]

(2) Let p > 1 be and assume J(x) ≥ J(y) if |x| ≤ |y|. Let T > 0, u0 ∈ Lp (Ω) and up,ε the unique solution of PpJε ,Ω (u0 ). Then, if vp is the unique solution of Pp (u0 ), lim sup kup,ε (t, ·) − vp (t, ·)kLp (Ω) = 0.

ε→0 t∈[0,T ]

(3) Let T > 0, u0 ∈ L2 (Ω) ∩ W 1,∞ (Ω) such that |∇u0 | ≤ 1 and consider u∞,ε the ε,Ω Ω unique solution of P∞ (u0 ). Then, if v∞ is the unique solution of P∞ (u0 ), we have lim sup ku∞,ε (t, ·) − v∞ (t, ·)kL2 (Ω) = 0.

ε→0 t∈[0,T ]

AMRTp-Lap

Part (2) was proved in [9], the other statements follows just by considering, as we did before, the Mosco convergence of the associated functionals. We leave the details to the reader. Example 6. In this case, let us also compute an explicit solution to the limit problem 1,Ω P∞ (u0 ) (to simplify we have considered ε = 1 in this example). Let us consider a recipient Ω = (0, l) with l an integer greater than 1, u0 = 0 and a source given by f (x, t) = χ[0,1] (x). Then the solution is given by u(x, t) = tχ[0,1] (x), for times t ∈ [0, 1]. For t ∈ [1, 3] we get  t−1   1 +   2   t−1 u(x, t) =   2     0

x ∈ [0, 1), x ∈ [1, 2), x∈ / [0, 2).

204


In general we have, until the recipient is full, for any  t − tk−1   k − 1 +   k      t − tk−1   k − 2 +   k  u(x, t) = ···       t − tk−1     k     0

k = 1, ..., l and for t ∈ [tk−1 , tk ) x ∈ [0, 1), x ∈ [1, 2)

x ∈ [k − 1, k), x∈ / [0, k)

Here tk = tk−1 + k is the first time when the solution reaches level k, that is u(tk , 0) = k. For times even greater, t ≥ tl , the solution turns out to be  t − tl   l+ x ∈ [0, 1),   l      t − tl   l−1+ x ∈ [1, 2), l u(x, t) =   ···        t − tl   1+ x ∈ [l − 1, l). l Hence, when the recipient is full the solution grows with speed 1/l uniformly in (0, l).

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